The Golden Ratio: 3-4-5

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 3−4−53-4-5 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 ABCABC with AB=5AB = 5, AC=4AC = 4, and BC=3BC = 3. Let OO be the foot of the angle bisector from BB. Draw a circle centered at OO with radius COCO. Then extend BOBO until it hits circle OO again at QQ. Let PP be the other point of intersection of these two curves. Then

PQBP=ϕ\frac{PQ}{BP} = \phi

Problem 1) Can you prove the above proposition?

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

Credit: Cut-the-Knot

Here is the next post.

#Geometry #Construction #CosinesGroup #Triangles #GoldenRatio

Note by Bob Krueger
7 years, 5 months ago

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Comments

For problem 2, simply reflect the entire figure about ACAC. 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.

Daniel Liu - 7 years, 5 months ago

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This is probably the easiest way to prove it.

Bob Krueger - 7 years, 5 months ago

Since 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.

Bob Krueger - 7 years, 5 months ago
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