Wednesday, January 21, 2015

Geometry Problem 1076: Square, 45 Degrees, Angle, Sum, Distance, Auxiliary Construction

Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Locate point H on the circle such that ∠ (HAF)= ∠ (DAF)= alpha
Triangles FAD congruence to HAF ( Case SAS) => FH=FD= b and ∠(AHF)=90
We have ∠ (EAH)=45- alpha=∠ (BAE) => triangles EAH congruence to EAB
So EH=EB=a and ∠ (EHA) 90
So x= a+b

2. Rotate anti-clockwise 90°, center at A.
Then D→B. Let F→G.

Then AF=AG and ∠EAF=∠EAG.
So ΔEAF congruent to ΔEAG.

Hence, x=EF=EG=a+b.

3. Fold AB, AD along AE, AF respectively.
B, D land at the same point P on EF.
So x = EP + PF = EB + DF = a + b.

4. Extend CD to G such that DG = a, then Tr.s ABE and ADG are congruent from which we can deduce that Tr.s AFE and AGF are also congruent.
Hence x= FG = a+b

Sumith Peiris
Moratuwa
Sri Lanka

5. if you rotate the 45 degree wedge, while extending and shortening AF and AE so they F and E are always on the square, EF will form a quarter circle, kinda like what is happening here: https://www.wikihow.com/images/thumb/c/c0/Draw-a-Parabolic-Curve-%28a-Curve-with-Straight-Lines%29-Step-18.jpg/aid5583818-v4-728px-Draw-a-Parabolic-Curve-%28a-Curve-with-Straight-Lines%29-Step-18.jpg

so EF will be tangent to a circle centered at A with radius AB. let the point of tangency be n. because of tangent properties, BE = a = EN and DF = b = FN. FN + EN = x, so a + b = x

6. See the drawing

Define α = ∠BAF and β = ∠DAE
α+β+45°=90° => α+β = 45°
Define E’ rotation of E anticlockwise with an angle of 90°
=>AE=AE’, β = ∠DAE = ∠BAE’
∠FAE’= α+β = 45°
ΔFAE’ is congruent to ΔFAE (SAS) => FE’=FE
Therefore a+b=x
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