Why isn't the side of the earth facing the sun completely illuminated by sunlight from the north to the south pole every day of the year? We all know that the earth rotates every 24 hours, that it is tilted 23.5 degrees, that it circles the sun every 365 days, and that it is located approximately 148,600,000 km (92,333,000 miles) from the Sun.

We also know that we have 5 major circles of latitude, from north to South: the Arctic circle at 66.5 degrees north, the Tropic of Cancer at 23.5 north; the Equator at 0: the Tropic of Capricorn at 23.5 south, and the Antarctic Circle at 66.5 south.

I have read previous answers regarding questions similar to mine but feel that the question(s) were perhaps misinterpreted as to meaning why isn't Earths side opposite the sun illuminated given the sun's size, so I want to be more specific.

Why isn't the side of the earth facing the sun completely illuminated by sunlight from the north to the south pole every day of the year?

Because if the The Earths diameter is 12,740 km (7,917 miles) and the Sun's is 1,392,700 km (865,370 miles), or about 109 times greater, it would seem logical that the sun's rays would provide sunlight on the face of the Earth pointed towards the Sun, completely from the North Pole to the South Pole, and yet we know that it doesn't...  Why so?
If the diameter of the Sun is $R_S$, the diameter of the Earth is $R_E$, and the orbital radius of the Earth is $r$, as in the drawing below (sorry about the quality),  then you can do some geometry and compute $\theta$, which is the angle at which the light from the edge of the Sun's disk as seen from Earth is tangent to the Earth's surface referenced to a line between the centres of the Sun and the Earth.



The expression for $\theta$ comes out to be

$$\theta = \sin^{-1}\left(\frac{R_S - R_E}{r}\right)$$

(You get this by extending the tangent line and the line connecting the centres, realising that this makes two similar right triangles with the angle at the tip being $\theta$, and the base of the larger one being $R_S$, then working out what the hypotenuse of the larger triangle is, which gives you $\sin\theta$.)

Since $\theta$ is small we can just say $\sin \theta = \theta$ in the usual way, and plugging in values we get $\theta \approx 0.0046\,\mathrm{rad}$.

If the angle at which the Earth's rotation axis is tilted is more than $\theta$, then there will be a period of the year during which the poles are not illuminated as the point at which the Sun's light is tangent to the Earth's surface is sunward of the pole.  Well, the angle of tilt is $23.5^\circ \approx 0.41\,\mathrm{rad}$, which is about $90\theta$.