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- October 27, 2018 at 11:46 am #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.

October 28, 2018 at 4:59 am #24000This is a very interesting problem. The

**easy**solution is using brute force.**Short Answer:**1.61 (approximately); that is the**Golden Ratio****Long Answer:**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.

October 28, 2018 at 5:09 am #24002I 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.

###### Attachments:

You must be logged in to view attached files.October 28, 2018 at 6:42 am #24016Here is a link to some sequential hints on this problem (I replaced some of the coordinate bashing by elementary geometry).

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