 # Geometry

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• #23999

Given a right triangle ABC with <ACB =90.  Let CH, H lying on AB, be the altitude to AB and P and Q be the tangent points of the incircle of triangle ABC to AC and BC. If AQ is perpendicular to HP, find the ratio of AH to BH.

#24000

This is a very interesting problem. The easy solution is using brute force.

Short Answer: 1.61 (approximately); that is the Golden Ratio

Suppose A= (0,a), C = (0,0), B = (b, 0).

Then AH /BH = $\frac {a^2} {b^2}$ (why? either use geometry or brute force computation using coordinate geometry).

The fact that HP is perpendicular to AQ provides with the following equation: $$1 + \frac{b^2}{a^2} = \frac{a^2}{b^2}$$

(How? Find inradius, find the coordinates of P and Q, compute slopes of HP and AQ, set the product of the slopes equal to -1)

Assume $\frac{a^2}{b^2} = t$ and solve for t.

I will upload the computations if you are unable to do it.

#24002

I attached the coordinate geometry computations anyway. But I strongly suggest that you try this on your own first. It is not hard. Use the steps indicated above.

If you are absolutely stuck then look into the attached pdf file.

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#24016

Here is a link to some sequential hints on this problem (I replaced some of the coordinate bashing by elementary geometry).

https://www.cheenta.com/golden-ratio-and-right-triangles/

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