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Triangle | |
A triangle | |
Edges and vertices | 3 |
Schläfli symbol | {3} (for equilateral) |
Area | various methods; see below |
Internal angle (degrees) | 60° (for equilateral) |
Triangle | |
A triangle | |
Edges and vertices | 3 |
Schläfli symbol | {3} (for equilateral) |
Area | various methods; see below |
Internal angle (degrees) | 60° (for equilateral) |
A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted .
In Euclidean geometry any three points, when non-collinear, determine a unique triangle and a unique plane (i.e. a two-dimensional Euclidean space).
Triangles can be classified according to the relative lengths of their sides:
Equilateral | Isosceles | Scalene |
Hatch marks, also called tick marks, are used in diagrams of triangles and other geometric figures to identify sides of equal lengths. A side can be marked with a pattern of "ticks", short line segments in the form of tally marks; two sides have equal lengths if they are both marked with the same pattern. In a triangle, the pattern is usually no more than 3 ticks. An equilateral triangle has the same pattern on all 3 sides, an isosceles triangle has the same pattern on just 2 sides, and a scalene triangle has different patterns on all sides since no sides are equal. Similarly, patterns of 1, 2, or 3 concentric arcs inside the angles are used to indicate equal angles. An equilateral triangle has the same pattern on all 3 angles, an isosceles triangle has the same pattern on just 2 angles, and a scalene triangle has different patterns on all angles since no angles are equal.
Triangles can also be classified according to their internal angles, measured here in degrees.
A triangle that has two angles with the same measure also has two sides with the same length, and therefore it is an isosceles triangle. It follows that in a triangle where all angles have the same measure, all three sides have the same length, and such a triangle is therefore equilateral.
Right | Obtuse | Acute |
Oblique |
Triangles are assumed to be two-dimensional plane figures, unless the context provides otherwise (see Non-planar triangles, below). In rigorous treatments, a triangle is therefore called a 2-simplex (see also Polytope). Elementary facts about triangles were presented by Euclid in books 1–4 of his Elements, around 300 BC.
The measures of the interior angles of a triangle in Euclidean space always add up to 180 degrees.[5] This allows determination of the measure of the third angle of any triangle given the measure of two angles. An exterior angle of a triangle is an angle that is a linear pair (and hence supplementary) to an interior angle. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. The sum of the measures of the three exterior angles (one for each vertex) of any triangle is 360 degrees.[note 2]
Two triangles are said to be similar if every angle of one triangle has the same measure as the corresponding angle in the other triangle. The corresponding sides of similar triangles have lengths that are in the same proportion, and this property is also sufficient to establish similarity.
Some basic theorems about similar triangles are:
Two triangles that are congruent have exactly the same size and shape:[note 4] all pairs of corresponding interior angles are equal in measure, and all pairs of corresponding sides have the same length. (This is a total of six equalities, but three are often sufficient to prove congruence.)
Some sufficient conditions for a pair of triangles to be congruent are:
An important condition is:
Using right triangles and the concept of similarity, the trigonometric functions sine and cosine can be defined. These are functions of an angle which are investigated in trigonometry.
A central theorem is the Pythagorean theorem, which states in any right triangle, the square of the length of the hypotenuse equals the sum of the squares of the lengths of the two other sides. If the hypotenuse has length c, and the legs have lengths a and b, then the theorem states that
The converse is true: if the lengths of the sides of a triangle satisfy the above equation, then the triangle has a right angle opposite side c.
Some other facts about right triangles:
For all triangles, angles and sides are related by the law of cosines and law of sines (also called the cosine rule and sine rule).
The triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side. That sum can equal the length of the third side only in the case of a degenerate triangle, one with collinear vertices. It is not possible for that sum to be less than the length of the third side.
Three positive angles α, β, and γ, each of them less than 180°, are the angles of a triangle if and only if any one of the following conditions holds:[6]
There are thousands of different constructions that find a special point associated with (and often inside) a triangle, satisfying some unique property: see the references section for a catalogue of them. Often they are constructed by finding three lines associated in a symmetrical way with the three sides (or vertices) and then proving that the three lines meet in a single point: an important tool for proving the existence of these is Ceva's theorem, which gives a criterion for determining when three such lines are concurrent. Similarly, lines associated with a triangle are often constructed by proving that three symmetrically constructed points are collinear: here Menelaus' theorem gives a useful general criterion. In this section just a few of the most commonly encountered constructions are explained.
A perpendicular bisector of a side of a triangle is a straight line passing through the midpoint of the side and being perpendicular to it, i.e. forming a right angle with it. The three perpendicular bisectors meet in a single point, the triangle's circumcenter, usually denoted by O; this point is the center of the circumcircle, the circle passing through all three vertices. The diameter of this circle, called the circumdiameter, can be found from the law of sines stated above. The circumcircle's radius is called the circumradius.
Thales' theorem implies that if the circumcenter is located on one side of the triangle, then the opposite angle is a right one. If the circumcenter is located inside the triangle, then the triangle is acute; if the circumcenter is located outside the triangle, then the triangle is obtuse.
An altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side. This opposite side is called the base of the altitude, and the point where the altitude intersects the base (or its extension) is called the foot of the altitude. The length of the altitude is the distance between the base and the vertex. The three altitudes intersect in a single point, called the orthocenter of the triangle, usually denoted by H. The orthocenter lies inside the triangle if and only if the triangle is acute.
An angle bisector of a triangle is a straight line through a vertex which cuts the corresponding angle in half. The three angle bisectors intersect in a single point, the incenter, usually denoted by I, the center of the triangle's incircle. The incircle is the circle which lies inside the triangle and touches all three sides. Its radius is called the inradius. There are three other important circles, the excircles; they lie outside the triangle and touch one side as well as the extensions of the other two. The centers of the in- and excircles form an orthocentric system.
A median of a triangle is a straight line through a vertex and the midpoint of the opposite side, and divides the triangle into two equal areas. The three medians intersect in a single point, the triangle's centroid or geometric barycenter, usually denoted by G. The centroid of a rigid triangular object (cut out of a thin sheet of uniform density) is also its center of mass: the object can be balanced on its centroid in a uniform gravitational field. The centroid cuts every median in the ratio 2:1, i.e. the distance between a vertex and the centroid is twice the distance between the centroid and the midpoint of the opposite side.
The midpoints of the three sides and the feet of the three altitudes all lie on a single circle, the triangle's nine-point circle. The remaining three points for which it is named are the midpoints of the portion of altitude between the vertices and the orthocenter. The radius of the nine-point circle is half that of the circumcircle. It touches the incircle (at the Feuerbach point) and the three excircles.
The centroid (yellow), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red point) all lie on a single line, known as Euler's line (red line). The center of the nine-point circle lies at the midpoint between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter.
The center of the incircle is not in general located on Euler's line.
If one reflects a median in the angle bisector that passes through the same vertex, one obtains a symmedian. The three symmedians intersect in a single point, the symmedian point of the triangle.
There are various standard methods for calculating the length of a side or the measure of an angle. Certain methods are suited to calculating values in a right-angled triangle; more complex methods may be required in other situations.
In right triangles, the trigonometric ratios of sine, cosine and tangent can be used to find unknown angles and the lengths of unknown sides. The sides of the triangle are known as follows:
The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. In our case
Note that this ratio does not depend on the particular right triangle chosen, as long as it contains the angle A, since all those triangles are similar.
The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. In our case
The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. In our case
The acronym "SOH-CAH-TOA" is a useful mnemonic for these ratios.
The inverse trigonometric functions can be used to calculate the internal angles for a right angled triangle with the length of any two sides.
Arcsin can be used to calculate an angle from the length of the opposite side and the length of the hypotenuse.
Arccos can be used to calculate an angle from the length of the adjacent side and the length of the hypontenuse.
Arctan can be used to calculate an angle from the length of the opposite side and the length of the adjacent side.
In introductory geometry and trigonometry courses, the notation sin−1, cos−1, etc., are often used in place of arcsin, arccos, etc. However, the arcsin, arccos, etc., notation is standard in higher mathematics where trigonometric functions are commonly raised to powers, as this avoids confusion between multiplicative inverse and compositional inverse.
The law of sines, or sine rule,[7] states that the ratio of the length of a side to the sine of its corresponding opposite angle is constant, that is
This ratio is equal to the diameter of the circumscribed circle of the given triangle. Another interpretation of this theorem is that every triangle with angles α, β and γ is similar to a triangle with side lengths equal to sin α, sin β and sin γ. This triangle can be constructed by first constructing a circle of diameter 1, and inscribing in it two of the angles of the triangle. The length of the sides of that triangle will be sin α, sin β and sin γ. The side whose length is sin α is opposite to the angle whose measure is α, etc.
The law of cosines, or cosine rule, connects the length of an unknown side of a triangle to the length of the other sides and the angle opposite to the unknown side.[7] As per the law:
For a triangle with length of sides a, b, c and angles of α, β, γ respectively, given two known lengths of a triangle a and b, and the angle between the two known sides γ (or the angle opposite to the unknown side c), to calculate the third side c, the following formula can be used:
If the lengths of all three sides of any triangle are known the three angles can be calculated:
The law of tangents or tangent rule, can be used to find a side or an angle when you know two sides and an angle or two angles and a side. It states that:[8]
"Solution of triangles" is the historical term for the solving of the main trigonometric problem: to find missing characteristics of a triangle (three angles, the lengths of the three sides etc.) when at least three of these characteristics are given. The triangle can be located on a plane or on a sphere. This problem often occurs in various trigonometric applications, such as geodesy, astronomy, construction, navigation etc.
Calculating the area T of a triangle is an elementary problem encountered often in many different situations. The best known and simplest formula is:
where b is the length of the base of the triangle, and h is the height or altitude of the triangle. The term "base" denotes any side, and "height" denotes the length of a perpendicular from the vertex opposite the side onto the line containing the side itself. In 499 CE Aryabhata, a great mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, used this method in the Aryabhatiya (section 2.6).
Although simple, this formula is only useful if the height can be readily found, which is not always the case. For example, the surveyor of a triangular field might find it relatively easy to measure the length of each side, but relatively difficult to construct a 'height'. Various methods may be used in practice, depending on what is known about the triangle. The following is a selection of frequently used formulae for the area of a triangle.[9]
The height of a triangle can be found through the application of trigonometry.
Knowing SAS: Using the labels in the image on the right, the altitude is h = a sin . Substituting this in the formula derived above, the area of the triangle can be expressed as:
(where α is the interior angle at A, β is the interior angle at B, is the interior angle at C and c is the line AB).
Furthermore, since sin α = sin (π − α) = sin (β + ), and similarly for the other two angles:
Knowing AAS:
and analogously if the known side is a or c.
Knowing ASA:[10]
and analogously if the known side is b or c.
The shape of the triangle is determined by the lengths of the sides. Therefore the area can also be derived from the lengths of the sides. By Heron's formula:
where is the semiperimeter, or half of the triangle's perimeter.
Three other equivalent ways of writing Heron's formula are
The area of a parallelogram embedded in a three-dimensional Euclidean space can be calculated using vectors. Let vectors AB and AC point respectively from A to B and from A to C. The area of parallelogram ABDC is then
which is the magnitude of the cross product of vectors AB and AC. The area of triangle ABC is half of this,
The area of triangle ABC can also be expressed in terms of dot products as follows:
In two-dimensional Euclidean space, expressing vector AB as a free vector in Cartesian space equal to (x1,y1) and AC as (x2,y2), this can be rewritten as:
If vertex A is located at the origin (0, 0) of a Cartesian coordinate system and the coordinates of the other two vertices are given by B = (xB, yB) and C = (xC, yC), then the area can be computed as 1⁄2 times the absolute value of the determinant
For three general vertices, the equation is:
which can be written as
If the points are labeled sequentially in the counterclockwise direction, the above determinant expressions are positive and the absolute value signs can be omitted.[11] The above formula is known as the shoelace formula or the surveyor's formula.
If we locate the vertices in the complex plane and denote them in counterclockwise sequence as a = xA + yAi, b = xB + yBi, and c = xC + yCi, and denote their complex conjugates as , , and , then the formula
is equivalent to the shoelace formula.
In three dimensions, the area of a general triangle A = (xA, yA, zA), B = (xB, yB, zB) and C = (xC, yC, zC) is the Pythagorean sum of the areas of the respective projections on the three principal planes (i.e. x = 0, y = 0 and z = 0):
The area within any closed curve, such as a triangle, is given by the line integral around the curve of the algebraic or signed distance of a point on the curve from an arbitrary oriented straight line L. Points to the right of L as oriented are taken to be at negative distance from L, while the weight for the integral is taken to be the component of arc length parallel to L rather than arc length itself.
This method is well suited to computation of the area of an arbitrary polygon. Taking L to be the x-axis, the line integral between consecutive vertices (xi,yi) and (xi+1,yi+1) is given by the base times the mean height, namely (xi+1 − xi)(yi + yi+1)/2. The sign of the area is an overall indicator of the direction of traversal, with negative area indicating counterclockwise traversal. The area of a triangle then falls out as the case of a polygon with three sides.
While the line integral method has in common with other coordinate-based methods the arbitrary choice of a coordinate system, unlike the others it makes no arbitrary choice of vertex of the triangle as origin or of side as base. Furthermore the choice of coordinate system defined by L commits to only two degrees of freedom rather than the usual three, since the weight is a local distance (e.g. xi+1 − xi in the above) whence the method does not require choosing an axis normal to L.
When working in polar coordinates it is not necessary to convert to cartesian coordinates to use line integration, since the line integral between consecutive vertices (ri,θi) and (ri+1,θi+1) of a polygon is given directly by riri+1sin(θi+1 − θi)/2. This is valid for all values of θ, with some decrease in numerical accuracy when |θ| is many orders of magnitude greater than π. With this formulation negative area indicates clockwise traversal, which should be kept in mind when mixing polar and cartesian coordinates. Just as the choice of y-axis (x = 0) is immaterial for line integration in cartesian coordinates, so is the choice of zero heading (θ = 0) immaterial here.
Three formulas have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides a, b, and c respectively as ma, mb, and mc and their semi-sum (ma + mb + mc)/2 as σ, we have[12]
Next, denoting the altitudes from sides a, b, and c respectively as ha, hb, and hc, and denoting the semi-sum of the reciprocals of the altitudes as we have[13]
And denoting the semi-sum of the angles' sines as S = [(sin α) + (sin β) + (sin γ)]/2, we have[14]
where D is the diameter of the circumcircle:
See Pick's theorem for a technique for finding the area of any arbitrary lattice polygon (one drawn on a grid with vertically and horizontally adjacent lattice points at equal distances, and with vertices on lattice points).
The theorem states:
where is the number of internal lattice points and B is the number of lattice points lying on the border of the polygon.
Numerous other area formulas exist, such as
where r is the inradius, and s is the semiperimeter (in fact this formula holds for all tangential polygons);
for circumdiameter D; and[16]
for angle α ≠ 90°.
Denoting the radius of the inscribed circle as r and the radii of the excircles as r1, r2, and r3, the area can be expressed as[17]
In 1885, Baker[18] gave a collection of over a hundred distinct area formulas for the triangle. These include:
for circumradius (radius of the circumcircle) R, and
The area of any triangle with perimeter p is less than or equal to with equality holding if and only if the triangle is equilateral.[19][20]:657
Other upper bounds on the area T are given by[21]:p.290
both again holding if and only if the triangle is equilateral.
There are infinitely many lines that bisect the area of a triangle.[22] Three of them are the medians, which are the only area bisectors that go through the centroid. Three other area bisectors are parallel to the triangle's sides.
Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter. There can be one, two, or three of these for any given triangle.
The formulas in this section are true for all Euclidean triangles.
The medians and the sides are related by[23]:p.70
and equivalently for mb and mc.
For angle α opposite side a, the length of the internal angle bisector is given by
for semiperimeter s, where the bisector length is measured from the vertex to where it meets the opposite side.
The interior perpendicular bisectors are given by
where the sides are and the area is [24]:Thm 2
The altitude from, for example, the side of length a is
The following formulas involve the circumradius R and the inradius r:
where ha etc. are the altitudes to the subscripted sides;[23]:p.79
Suppose two adjacent but non-overlapping triangles share the same side of length f and share the same circumcircle, so that the side of length f is a chord of the circumcircle and the triangles have side lengths (a, b, f) and (c, d, f), with the two triangles together forming a cyclic quadrilateral with side lengths in sequence (a, b, c, d). Then[26]:84
Let M be the centroid of a triangle with vertices A, B, and C, and let P be any interior point. Then the distances between the points are related by[26]:174
Let pa, pb, and pc be the distances from the centroid to the sides of lengths a, b, and c. Then[26]:173
The product of two sides of a triangle equals the altitude to the third side times the diameter of the circumcircle.[23]:p.64
Carnot's Theorem states that the sum of the distances from the circumcenter to the three sides equals the sum of the circumradius and the inradius.[23]:p.83 Here a segment's length is considered to be negative if and only if the segment lies entirely outside the triangle. This method is especially useful for deducing the properties of more abstract forms of triangles, such as the ones induced by Lie algebras, that otherwise have the same properties as usual triangles.
Euler's theorem states that the distance d between the circumcenter and the incenter is given by[23]:p.85
or equivalently
where R is the circumradius and r is the inradius. Thus for all triangles R ≥ 2r, with equality holding for equilateral triangles.
If we denote that the orthocenter divides one altitude into segments of lengths u and v, another altitude into segment lengths w and x, and the third altitude into segment lengths y and z, then uv = wx = yz.[23]:p.94
The distance from a side to the circumcenter equals half the distance from the opposite vertex to the orthocenter.[23]:p.99
The sum of the squares of the distances from the vertices to the orthocenter plus the sum of the squares of the sides equals twelve times the square of the circumradius.[23]:p.102
Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the Morley triangle.
As discussed above, every triangle has a unique inscribed circle (incircle) that is interior to the triangle and tangent to all three sides.
Every triangle has a unique Steiner inellipse which is interior to the triangle and tangent at the midpoints of the sides. Marden's theorem shows how to find the foci of this ellipse.[27] This ellipse has the greatest area of any ellipse tangent to all three sides of the triangle.
The Mandart inellipse of a triangle is the ellipse inscribed within the triangle tangent to its sides at the contact points of its excircles.
For any ellipse inscribed in a triangle ABC, let the foci be P and Q. Then[28]
The Lemoine hexagon is a cyclic hexagon with vertices given by the six intersections of the sides of a triangle with the three lines that are parallel to the sides and that pass through its symmedian point. In either its simple form or its self-intersecting form, the Lemoine hexagon is interior to the triangle with two vertices on each side of the triangle.
Every acute triangle has three inscribed squares (squares in its interior such that all four of a square's vertices lie on a side of the triangle, so two of them lie on the same side and hence one side of the square coincides with part of a side of the triangle). In a right triangle two of the squares coincide and have a vertex at the triangle's right angle, so a right triangle has only two distinct inscribed squares. An obtuse triangle has only one inscribed square, with a side coinciding with part of the triangle's longest side. Within a given triangle, a longer common side is associated with a smaller inscribed square. If an inscribed square has side of length q and the triangle has a side of length a, part of which side coincides with a side of the square, then q, a, and the triangle's area T are related according to[29]
The largest possible ratio of the area of the inscribed square to the area of the triangle is 1/2, which occurs when a2 = 2T, q = a/2, and the altitude of the triangle from the base of length a is equal to a. The smallest possible ratio of the side of one inscribed square to the side of another in the same non-obtuse triangle is [30]
From an interior point in a reference triangle, the nearest points on the three sides serve as the vertices of the pedal triangle of that point. If the interior point is the circumcenter of the reference triangle, the vertices of the pedal triangle are the midpoints of the reference triangle's sides, and so the pedal triangle is called the midpoint triangle or medial triangle. The midpoint triangle subdivides the reference triangle into four congruent triangles which are similar to the reference triangle.
The Gergonne triangle or intouch triangle of a reference triangle has its vertices at the three points of tangency of the reference triangle's sides with its incircle. The extouch triangle of a reference triangle has its vertices at the points of tangency of the reference triangle's excircles with its sides (not extended).
As mentioned above, every triangle has a unique circumcircle, a circle passing through all three vertices, whose center is the intersection of the perpendicular bisectors of the triangle's sides.
Further, every triangle has a unique Steiner circumellipse, which passes through the triangle's vertices and has its center at the triangle's centroid. Of all ellipses going through the triangle's vertices, it has the smallest area.
The Kiepert hyperbola is the unique conic which passes through the triangle's three vertices, its centroid, and its circumcenter.
One way to identify locations of points in (or outside) a triangle is to place the triangle in an arbitrary location and orientation in the Cartesian plane, and to use Cartesian coordinates. While convenient for many purposes, this approach has the disadvantage of all points' coordinate values being dependent on the arbitrary placement in the plane.
Two systems avoid that feature, so that the coordinates of a point are not affected by moving the triangle, rotating it, or reflecting it as in a mirror, any of which give a congruent triangle, or even by rescaling it to give a similar triangle. Trilinear coordinates specify the relative distances of a point from the sides, so that coordinates x : y : z indicate that the ratio of the distance of the point from the first side to its distance from the second side is x : y, etc. Barycentric coordinates of the form specify the point's location by the relative weights that would have to be put on the three vertices in order to balance the otherwise weightless triangle on the given point.
A non-planar triangle is a triangle which is not contained in a (flat) plane. Some examples of non-planar triangles in non-Euclidean geometries are spherical triangles in spherical geometry and hyperbolic triangles in hyperbolic geometry.
While the measures of the internal angles in planar triangles always sum to 180°, a hyperbolic triangle has measures of angles that sum to less than 180°, and a spherical triangle has measures of angles that sum to more than 180°. A hyperbolic triangle can be obtained by drawing on a negatively curved surface, such as a saddle surface, and a spherical triangle can be obtained by drawing on a positively curved surface such as a sphere. Thus, if one draws a giant triangle on the surface of the Earth, one will find that the sum of the measures of its angles is greater than 180°; in fact it will be between 180° and 540°.[31] In particular it is possible to draw a triangle on a sphere such that the measure of each of its internal angles is equal to 90°, adding up to a total of 270°.
Specifically, on a sphere the sum of the angles of a triangle is
where f is the fraction of the sphere's area which is enclosed by the triangle. For example, suppose that we draw a triangle on the Earth's surface with vertices at the North Pole, at a point on the equator at 0° longitude, and a point on the equator at 90° West longitude. The great circle line between the latter two points is the equator, and the great circle line between either of those points and the North Pole is a line of longitude; so there are right angles at the two points on the equator. Moreover, the angle at the North Pole is also 90° because the other two vertices differ by 90° of longitude. So the sum of the angles in this triangle is 90° + 90° + 90° = 270°. The triangle encloses 1/4 of the northern hemisphere (90°/360° as viewed from the North Pole) and therefore 1/8 of the Earth's surface, so in the formula f = 1/8; thus the formula correctly gives the sum of the triangle's angles as 270°.
From the above angle sum formula we can also see that the Earth's surface is locally flat: If we draw an arbitrarily small triangle in the neighborhood of one point on the Earth's surface, the fraction f of the Earth's surface which is enclosed by the triangle will be arbitrarily close to zero. In this case the angle sum formula simplifies to 180°, which we know is what Euclidean geometry tells us for triangles on a flat surface.
Rectangles have been the most popular and common geometric form for buildings since the shape is easy to stack and organize; as a standard, it is easy to design furniture and fixtures to fit inside rectangularly shaped buildings. But triangles, while more difficult to use conceptually, provide a great deal of strength. As computer technology helps architects design creative new buildings, triangular shapes are becoming increasingly prevalent as parts of buildings and as the primary shape for some types of skyscrapers as well as building materials. In Tokyo in 1989, architects had wondered whether it was possible to build a 500-story tower to provide affordable office space for this densely packed city, but with the danger to buildings from earthquakes, architects considered that a triangular shape would have been necessary if such a building was ever to have been built (it hasn't by 2011).[32]
In New York City, as Broadway crisscrosses major avenues, the resulting blocks are cut like triangles, and buildings have been built on these shapes; one such building is the triangularly shaped Flatiron Building which real estate people admit has a "warren of awkward spaces that do not easily accommodate modern office furniture" but that has not prevented the structure from becoming a landmark icon.[33] Designers have made houses in Norway using triangular themes.[34] Triangle shapes have appeared in churches[35] as well as public buildings including colleges[36] as well as supports for innovative home designs.[37]
Triangles are sturdy; while a rectangle can collapse into a parallelogram from pressure to one of its points, triangles have a natural strength which supports structures against lateral pressures. A triangle will not change shape unless its sides are bent or extended or broken or if its joints break; in essence, each of the three sides supports the other two. A rectangle, in contrast, is more dependent on the strength of its joints in a structural sense. Some innovative designers have proposed making bricks not out of rectangles, but with triangular shapes which can be combined in three dimensions.[38] It is likely that triangles will be used increasingly in new ways as architecture increases in complexity. It is important to remember that triangles are strong in terms of rigidity, but while packed in a tessellating arrangement triangles are not as strong as hexagons under compression (hence the prevalence of hexagonal forms in nature). Tessellated triangles still maintain superior strength for cantilevering however, and this is the basis for one of the strongest man made structures, the tetrahedral truss.
A construction company said Thursday that it has designed a 500-story skyscraper for Tokyo, ... The building is shaped like a triangle, becoming smaller at the top to help it absorb shock waves. It would have a number of tunnels to let typhoon winds pass through rather than hitting the building with full force.
Though it is hard to configure office space in a triangle
Local zoning restrictions determined both the plan and the height of the Triangle House in Nesodden, Norway, which offers views toward the sea through a surrounding pine forest.
the classical functions of a church in two pure forms: a stark triangle of glass and, inside it, a rounded, egglike structure made of wood.
More strength, less material ... They share a common material language of structural steel, glass and metal panels, and stucco cladding; their angular, dynamic volumes, folded roof plates, and triangular forms are meant to suggest the plate tectonics of the shifting ground planes they sit on.
Perched like a tree house, the $300,000 structure sits lightly on the terrain, letting the land flow beneath it. Much of the building rests on three triangular heavy-timber frames on a concrete pad.
Bricks are among the world's oldest building materials — the first were used as long ago as 7,500 B.C. ... An especially beautiful proposal by Rizal Muslimin at the Massachusetts Institute of Technology came in as a runner-up: BeadBricks are flat, triangular bricks that can be combined in three dimensions (rather than the usual two).
Wikimedia Commons has media related to Triangles. |
Look up triangle in Wiktionary, the free dictionary. |


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Fantasy Future Rankings: Week 8
Is there reason to be concerned with the production of Adrian Peterson?
Stop using past performance to evaluate fantasy players. Start using the future. Each week, numberFire will provide updated projections for future fantasy points through Week 17. To go with the numbers, Field Yates will provide analysis on what the numbers reveal. Fantasy players have a new resource to evaluate trades and roster moves: The Fantasy Future Rankings.
Adrian Peterson was the undisputed most valuable player in fantasy football in 2012. That honor (although unofficial) dovetailed nicely with the literal hardware he received as the league's "real-life" Most Valuable Player.
He was the unquestioned top draft choice entering this season as well, leading the way among a field of running backs that offered an intriguing mix of proven veterans (Arian Foster, Marshawn Lynch); upstarts with potential (Doug Martin, Trent Richardson); and young-and-proven backs in new systems (LeSean McCoy, Jamaal Charles).
Few decisions required less brainwork than snatching Peterson with the first overall pick if you owned it in your re-draft league.
If you picked Peterson, you probably aren't regretting the decision, as he's been among the best running backs in terms of overall scoring (No. 7) and even better in per-game average, as he has only played six games but still leads Foster, Martin and Richardson, among other notable backs.
But many of Peterson's owners weren't picking him for very good results; they were picking Peterson with a duplication of his dominant -- perhaps transcendent -- 2012 season in mind. So far, it hasn't happened.
To see the rest of the fantasy future rankings and which players have the most value to the end of the season, you must be an ESPN Insider.
Become an insider and get access to all our articles and tools.
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Robert Griffin III is beginning to look like his dynamic self again. How will he fare against the Broncos this week? Rankings »
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text: TOP 10 Top 10 Better Man Top 10: Ocean Rivalries Back to Top ▲ Top 10: Ocean Rivalries 3 Squid vs. Sperm whale Share on Facebook Share on Twitter 0 Shares Arguably the most famous of all the great ocean rivalries, a sperm whale against a squid is the stuff of legend. Sperm whales have the greatest diving capacity of any cetacean, and can reach depths of up to 9,842 feet. At these depths, they find their favorite food, giant squids. Giant squids can reach up to 60 feet in length, and weigh up to one ton, which might sound like a lot but they're dwarfed by the 50 tons of the sperm whale. What results is usually an epic struggle in the deep, with the sperm whale often (though not always) getting its meal. Even then, the squid will struggle valiantly, using its sharp beak to inflict huge gashes on the whale. Next Page TOP 10 9 8 7 6 5 4 3 2 1 Comments Show Comments More Like This Promoted Around the Web Special Features


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April 28, 2014 -- A Cinco de Mayo Fiesta sponsored by the Spanish Department will take place in Northwestern Oklahoma State University's Student Center Ballroom in Alva Thursday, May 1, and is open to the public.The event will kick off from 3:30 to 4:30 p.m. with a Salsa Tasting Contest followed by an authentic Mexican food meal at 5 p.m. All proceeds from the evening's events will go toward the Spanish Scholarship Fund.
April 26, 2013 -- Watching traditional Mexican folk dancers, eating authentic Mexican food and making salsa for a salsa contest are how Northwestern Oklahoma State University students will celebrate Cinco de Mayo. Northwestern's celebration is planned for May 2.
On April 29, the Peninsular Spanish American Culture class at Northwestern Oklahoma State University will play host to a fundraiser in celebration of Cinco de Mayo, with its main goal to raise money for the Spanish Scholarship Fund. Lunch will be served from 12:30 to 1:30 p.m.


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Rook knocked on the front door of the Tennyson Household, as he had come to keep an eye on Jen, as Carl and Sandra, Jen's Parents, were currently at work during the summer. The Task of making sure Jen SURVIVED was left to her dear sweet friends.
Rook had been asked to come see Jen who, according to Magister Tennyson, was confined to her bed by order of the Plumbers, this was instigated due to knowledge of the LAST time Jen had tried to use her Aliens while Sick...well, Rook wasn't sure WHAT happened. But it involved Hives, Ice Fire, and A Swarm of cockroaches and WAY too many Alien Fluids...according to Gwen Tennyson.
A frazzled looking Kevin Levin opened the door, sighing in relief and waving Rook in, walking toward the kitchen as rook followed as small package in hand.
"Hello, Kevin. How is she?" Rook fought the enforced manners which decreed he call Kevin, Mr. Levin, but Kevin didn't appreciate it, so manners dictated he call him by his first name.
"Nothing two aspirin and a cattle prod wouldn't cure." Kevin dryly quipped.
"KEVIN!!" A Voice snapped from the kitchen.
The Two men stepped inside to see Gwen Tennyson, a knife in hand and poised over a piece of toast that had half of its crust cleanly sliced off. The Red-head's eyes were trained on Kevin in a deep frown.
"If it's the Truth, you should just say it, no whacked out lies." Kevin Quoted pointing at Gwen. "That's what YOU always say."
"I also say, 'Stop Tormenting My Cousin or I'll Kick you in places that you should NEVER kick a man." Gwen said the knife aimed at Kevin in warning, who responded with a nervous gulp and shuffling on his feet.
"What places are those? "Rook asked innocently.
Gwen and Kevin turned to stare at him briefly, before turning to look at each other and shaking their heads.
"You'll know when you've stayed on Earth a little longer." Gwen explained, before placing the crust-less toast onto a small plate, before handing it and a Glass of Orange juice to Rook.
"People keep saying that," Rook remarked curiously, accepting the glass in one hand as he balanced the tray on the small box in the crook of his other arm.
"Well, Jen's is upstairs lying in her bed." Gwen said, grabbing her purse hanging from hook on the wall. "She's been positively miserable since the cold first set in."
"I hate to think how miserable she'll be when she gets pregnant." Kevin shudders in fear. "Although not much chance of that happening if she keeps throwing food on men that get close to her bed."
Kevin remarked sourly, earning a confused look from Rook and a stink eye from Gwen.
"Well, what Did you expect?" Scolded Gwen. "I told you not to give her that jelly toast with the Crust on!"
"So, she smacks it onto the front of my new shirt, Jelly-Side Up?" Kevin retorted.
"She's sick, Kevin!" Gwen's voice rose, only to quickly look up the stairs in worry, before lowering her voice and pointing at Kevin.
"Look, we have to go anyway look in on that case of Sonic Wave's occurring in Pedmic 30 miles outside of Bellwood anyway." Gwen turns to Rook.
"Rook, you're sure you can look after Jen while we're gone?" Gwen inquired to the Alien Plumber Agent.
"I have dealt with sick people before, Gwen Tennyson." Rook stated formally as always. Gwen gave him a bemused look.
"Not Jen when she's sick." She stated simply.
"Come on, let's go before the Dragon Lady wales up and decides she wants US for Lunch." Kevin said in a hurried tone.
Gwen rolled her eyes, but waved bye to Rook, informing him where the food, medicine and drinks for Jen were set aside before heading out the door with Kevin.
Rook knew the way to Jen's bedroom, supervised of course by MR. TENNYSON, of course, and had spent some time in there waiting for her to finish homework to go on a mission or to just play that "Sumo Slammer's" game she adored so much.
Stepping into the room he noticed RIGHT away how badly Jen was sick, for instance....the room was a disaster!
Of course, Jen was hardly the SPICK and SPAN Girl that her Cousin Gwen was, thou what Spick and Span had to do with clean living domiciles was beyond Rook. He just took the Earth Idiom's as they came.
But, the room was kept moderately tidy, the bed was usually made and clothes were folded and put away.
But now....the room was littered with tissues, empty cartons of orange juice, with the orange juice stained glasses on the nightstand beside her bed and the bed...well, the Bed Resembled a Lepidopterran Cocoon at the moment. The covers on the bed being all swaddled and wrapped up tightly around the bed occupant.
Rook knew there was someone in there, as he heard a loud cough and groan of misery from within, a whine he was FAIRLY familiar with for a while now.
"Jen?" Rook asked carefully, skipping to avoid some cartons and set down the package to brush clean the nightstand with a cloth stored in his belt, as well as putting down some of the glasses to clean at a later time.
The Blankets quivered as a head poked out of the Blanket Cocoon, it LOOKED like Jen Tennyson, but her face was flushed, her hair was sticking out (More than Normally)and her eyes were rimmed with red circles.
"Aww. Ma, can't I sleep in a b-URRK!!" Jen startled her head jerking back upon seeing Rook, who jumped a little by the sudden action.
"Ma! Your all Purple!!" Jen said in hushed shock.
"Jen, it's Rook. Your Boyfriend and Sharer of your flame and Vice Versa?" Rook explained to the groggy Shapeshifter.
"I don't know..." Jen muttered, pulling one arm free to pull her way out of the tangle she was caught inside of. "My Mom has a habit of spouting Hippie junk like That as well....AGGH!!"
Jen had gotten her arm free, only to slip halfway off the bed upside down, thankfully caught by Rook, who pulled her back up into a seated position.
Once seated on the bed and blankets were smoothed out, Jen grabbed hold of Rook's biceps to steady herself, the blood rushing to her head from the movement dizzying her badly.
"Oh..." Jen gasped, squeezing Rook's bicep's experimentally. "Yep. Either Mom's new, Oak-Grass Shake Diet is working or it really IS you Rook."
"Well, why wouldn't it be me?" Teased Rook slightly. "I was Rook when I went to bed last night."
"What Day is this?" Groaned Jen, rubbing her forehead as she attempted to lean back against the headboard. Rook grabbing the pillows and placing them behind her back and head respectively to give her comfort.
'The last thing I can remember is my mom giving me some Nyquil to help me sleep.' Jen thought.
"I believe it's Saturday, Amar." Rook mentioned, fluffing the pillow she was leaning on before settign about picking up the tissues from the floor and tossing them into the garbage can.
"Aaagggh." Jen his-growled in frustration. "I've been Sick Three Days and still feel like crap?"
Rook paused in his clean up to look up startled.
"How would you KNOW what "Feces" feel's like?" Rook queried.
"Mmmm. It's a expression." Jen told him.
"A very ODD and Unappealing one I might add."
"Rook, why are you here!?" Jen said, snapping her head in his direction, wincing at the pain from the sudden gesture.
"To keep an eye on you and make sure you well." Rook replied.
"Well" Sneered Jen. "You've come by, made sure, so just leave me some water, a couple boxes of tissues and enough Nyquil to kill a elephant and I'll be fine."
She huffed, turning to fold her arms, frowning in discomfort and agitated beyond belief.
Rook wasn't bothered by her Verbal lashing, truth be told he had been forewarned and prepared for this kinda behavior from Jen. As Magister Tennyson told him, Jen was a NASTY character when she was sick, but Rook had dealt with Vicious criminals A LOT since his joining the Plumbers. So Jen's moody, attitude wasn't gonna bother him none.
"I'm afraid I can't do that." Rook said pleasantly, once he had the floor clean and the tissues shoved into the garbage can, he rose to give her a warm smile.
"It's my right and Duty." Rook explained.
"As a plumber?" Jen mumbled. She jumped slightly when Rook kissed her warm cheek, the cool contrast of his lips startling her, as Rook placed his hand over her's and squeezed it gently.
"As your Boyfriend, and Flame Sharer." Rook whispered.
Turning he headed out the bedroom, Jen watched him leave and felt herself feeling even more Morose. And it Wasn't cause of the cold.


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Stoik Stitch Creator User Reviews
"basic problems with software"
Version: Stoik Stitch Creator 4.0.2822
I suppose there's a pro in it somewhere...
The file became corrupted. In the edit mode, I couldn't see the pattern because the gridlines were out of whack with each other - overlapping, etc. The problem kept getting worse, and after 6 days, the company has not replied. Also, it's difficult to choose a palate and color of thread. And, there's no ability to copy between projects. Very frustrating. Also, no direct way to move a part of the design.
"Easy to use"
Version: Stoik Stitch Creator 4.0
there are plenty of images you can use or you can use your own images or pictures.
you can personalise your cross stiching to suit who ever wants it.
you can make your family photo's into the template if you wish to make gifts out of.Cons
didnt find any
Great little programme........TRY IT....
"Is there anyone tried it before?"
Version: Stoik Stitch Creator 4.0
I do like the graphics, it looks like a real cross stitch. The software is user friendly, easy to use, offcourse it could be better. I downloaded it to my home and work computers (don't tell to my boss:) I am so excited to start stitching my model.
However even they are the same program I couldn't open the project, which I did at home, from work. At home it gave me almost the real colors of the picture but at work for the same project it shows the red and yellow areas as the shades of brown.
One didn't give the real colors while the same other did. I couldn't open the project from other computer even they are the same software. I need an explenation about these issues before deciding to buy it or not?? Otherwise the graphics are good and this software is easy to use except the part of deleting backstitch.
"So far it has quickly accomplished what I wanted."
Version: Stoik Stitch Creator 3
I like the length of time available to try PM Stitch Creator. I am not a designer nor artist; the learning curve was not too bad.
It took me some time to figure out how to use an existing picture and have the program turn it into counted cross stitch for me. But I did figure it out!
I've only been using the demo version but am impressed compared to what I have read about other software. I have used lots of quilting software and this, to me, was easier. My requirements were small - just wanted to have a picture (graphic) turned magically into a counted cross stitch pattern I could print or see well on the monitor. PM Stitch Creator did just that for me. When I can find the money, I will be purchasing the program.
Version: Stoik Stitch Creator 3
make you own designs from your favourite photos and photos of your favourite popband or whatever you want or would like to stitch from any kind of Photo
you can't create your own design from your head i thought you could input you own stitches or can ya?
"easy to use"
Version: Stoik Stitch Creator 3
i love to make my own patterns and the creator is easy to use
"Very impressive! Especially it's ability to transform photos."
Version: Stoik Stitch Creator 3
I take a lot of photographs and wanted to find a program that would transform the photos into cross stitch with out loosing the pictures quality. I was very impressed with the fact that you can choose the amount of colors, stiches and size to achieve a more detailed piece of work. I am happy to have been able to try this software. I will definitly buy it.
"I had problems"
Version: Stoik Stitch Creator 3
Instead of using a sketch book or Microsoft Paint, I could gather all the parts of my design in an easier way
I'm pretty tech savvy, but I found this program hard to use and not able to do what I wanted it to. When I scaned in designs, it did not accurately incorporate them into a design. I had to manually fix the problems, something that took hours. I tried every option available to get it more accurate, and it never got it better.
When I did start with a blank page and hand drew (clicking) the design, it was not easy to make changes or edit it. If a line was off center, there was no way to drag and drop it into the right place. The program said there was, but after following the help guide, it still didn't work like it said it should. The only way to fix these mistakes was to erase the line and start over in the correct place.
Although it did have the option of importing my colors using the DMC numbers, the colors shown on the screen weren't always good matches to what I had in front of me. Also if I added a color and then decided not to use it, I couldn't find any way to delete the color from the palette. That left me trying to remember which color I ended up using because there were too many to choose from.
Even once I got a design I liked, printing it was very hard. I wanted to remove the extra length and width of the design place, and it would not shrink it without shrinking (and distorting) the design. In order to print the design enlarged to fit the page, I had to use the "print screen" option to move it to Microsoft Word and then crop and enlarge the page to fill the screen.
Overall this program did not save time over the draw-and-erase-in-the-sketch-book method I've used until now. The only benefits was being able to save it (in Word) to print later copies, and having a pattern I could get all on one page. Those advantages aren't worth all the time it took me to use the program. I spent 6-7 hours to get a 3"x5" design.
I did try the advanced mode, and it didn't work any better. Maybe I'm trying to do things too advanced for the program. Overall this is not what I was looking for.
"Great tool at a good price"
Version: Stoik Stitch Creator 3
I am using PM Stitch Creator 3 to design rugs (hooked and braided). I was able to use the software in no-time flat.
"Embroidery Machines?"
Version: Stoik Stitch Creator 3
Really easy to use. Nice interface.
It would be nice if it had a plugin to digitalize for embroidery machines.
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Like the title asks, how often do you replace recoil springs? Do you recommend replacing the spring when buying an older used gun?
This is a discussion on How often do you replace recoil springs? within the General Firearm Discussion forums, part of the Related Topics category; Like the title asks, how often do you replace recoil springs? Do you recommend replacing the spring when buying an older used gun?...
Like the title asks, how often do you replace recoil springs? Do you recommend replacing the spring when buying an older used gun?
ἐν τούτῳ νίκα - In hoc signo vinces
Confidence- thrives on honesty, on honor, on the sacredness of obligations, on faithful protection and on unselfish performance. Without them it cannot live.-FDR
Good question, I'd like to know as well. I haven't replaced them in any of mine to include the p85 I bought used and carry. Although I've put at least 800 rounds through it with no issue so I don't think I need one. Anyone I'm interested in an answer to this too.
Watching for responses. I'm curious, too.
"If you make something idiot proof, someone will make a better idiot."
- Anon
depends on the gun
different manufactuerers have differing reccomendations.
Sig Sauer recommends replacing the recoil spring in the P238 every 1,500 rounds. It's listed in the manual that comes with the gun.
I have a little range book I use to keep track of rounds shot since last replacement. The springs are very cheap, so changing them out is no big deal.
"...people who carry a gun understand that they are arming themselves against a very unlikely event... People who arm themselves are not confused about the odds. They are concerned about the stakes. -Kathy Jackson
NRA Life Member
Depends on the gun and its intended use. For range guns, whenever I start having spring-related function issues (slide cycling), and for SD pistols every 2-3K rounds (except an STI Ranger with RecoilMaster). When buying a used gun, yes, I like to change out FP and recoil springs.
Chose a weapon that goes bang EVERY time!
You need to replace them? I replaced the one in my CZ-82 because I didn't know how used it was or what care it had been given and it ejected cases out of visual range. My used 4566 and 6906 worked so well I haven't replaced them. The Glock is OEM.
Retired USAF E-8. Lighten up and enjoy life because:
Paranoia strikes deep, into your life it will creep. It starts when you're always afraid... Buffalo Springfield - For What It's Worth
I have a 20 year old Glock 21 and I've never replaced the springs.
I never really thought about it.
Psalm 144:1
Blessed be the Lord, my rock, who trains my hands for war, and my fingers for battle.
Romans 13:4
But if you do wrong, be afraid, for he does not bear the sword in vain. For he is the servant of God, an avenger who carries out God's wrath on the wrongdoer.
+1 Moby on my 22 year old Glock. I do have some if they start to cause issues though.
Praise the Lord my Rock, who trains my hands for war, my fingers for battle --- Psalm 144
NRA Endowment Life
There are NO Silver Medals for Street Combat
Blue Thunder, I smell Victory in the Morning!
Of all of my guns I've only replaced one recoil spring and that was on an M1 carbine that probably had the same spring on it since 1945 and started to have ejection problems after I put about 2,000 rounds through it.
Now it ejects spent cartridges about 6 ft above my head and 8 feet behind me
A real man loves his wife, and places his family as the most important thing in life. Nothing has brought me more peace and content in life than simply being a good husband and father.
I replaced my glock originals with springs on stainless guides because I just didn't like the plastic ones. I expect that hundreds of thousands of cycles before they reach their fatigue life limit. That said, one could break and so you should keep a spare on hand. The part is cheap and it is easy to change. If you ever start to have problems with cycling the slide, swap out the recoil spring to see if it is the problem.
Yoda, I am, yes.
In 60 years of shooting I have never change a magazine spring and have changed only one recoil spring when I upgraded to a threaded barrel. I have several target pistols that have at least 20K rounds thru them, over the past 20 years, using the same magazines and springs since new without a magazine failure.
US Army 1953-1977
We, the People are the rightful masters of both Congress and the courts not to overthrow the Constitution, but to overthrow men who pervert the Constitution.
Abraham Lincoln
Compact and short barreled semi-autos should have their springs replaced more often.
Once your semi-automatic starts tossing your ejected brass out further than has been normal - it is time for a recoil spring change.
Spent cartridges being ejected out of the port further out than usual is indicative of the fact that the slide is moving to the rear with more speed and less resistance due to a weakened recoil spring.
If you are shopping around for a replacement recoil spring buy a Chrome Silicon recoil spring if possible. The have a useful spring life that is AT LEAST double that of the more common Carbon or Stainless steel spring.
Liberty Over Tyranny Μολὼν λαβέ
Every gun is different.
IIRC, on my old Browning BDM 9mm, I typically swapped to new springs at least every 5K rds. On the CZ P-01 9mm, every few thousand. On the "mouse" guns, every 750-1000 rds or so. Depends somewhat on how "hot" the rounds are that I've been feeding it, too. I've found that a steady diet of hot 124gr +P's on the Browning BDM meant things began feeling loosened-up earlier than otherwise, so recoil spring swaps made more sense to do on a faster schedule.
Your best weapon is your brain. Don't leave home without it.
Thoughts: Justifiable self defense (A.O.J.).
Explain: How does disarming victims reduce the number of victims?
Reason over Force: Why the Gun is Civilization (Marko Kloos).


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