Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.

Since we have already talked about several Golden figures, we are going to delve more deeply into where exactly we can find the golden ratio in other figures. Today we will talk about the ubiquitous \(3-4-5\) triangle. Everyone has learned that this is the simplest of all integral right triangles. But interestingly enough, it also contains the golden ratio. Here is the construction:

Above is the triangle \(ABC\) with \(AB = 5\), \(AC = 4\), and \(BC = 3\). Let \(O\) be the foot of the angle bisector from \(B\). Draw a circle centered at \(O\) with radius \(CO\). Then extend \(BO\) until it hits circle \(O\) again at \(Q\). Let \(P\) be the other point of intersection of these two curves. Then

\[\frac{PQ}{BP} = \phi\]

Problem 1) Can you prove the above proposition?

Problem 2) Extra: In the figure above, there is another point \(C'\). Prove that the circle \(O\) intersects the line segment \(AB\) at only one point \(C'\). Also, why might it be called \(C'\)?

Credit: Cut-the-Knot

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## Comments

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TopNewestSince no one has answered the first problem, I will assume that it was either too easy or too hard. As a hint for those who thought it was hard, try using the half angle formula for tangent, but do not calculate the angles directly.

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For problem 2, simply reflect the entire figure about \(AC\). You will find the circle in the center is the incircle from an incircle's definition of being the circle with center being the intersection of the angle bisectors, and the result follows.

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This is probably the easiest way to prove it.

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