I recently had fun playing around with the following problem, based on the last question of the 2003 AIME I.

In triangle let be the midpoint of , and let be the point on such that bisects angle . Let be the point on such that and are perpendicular. If meets at , find the ratio in terms of the side lengths of .

The interested reader might like to have an attempt at this problem before reading further. 🙂

Initially I solved this using vector geometry applying facts such as , and (see below). After seeing it led to a relatively simple answer I started looking for a more elegant approach and after seeing this page that shows methods using mass point geometry and applications of Menelaus’ theorem, came up with the following nice solution.

Extend to meet in as shown below. As is an angle bisector perpendicular to , reflects in line to so is the midpoint of .

Also the lines and are both perspective from the point , so the cross ratios and are equal (indeed, by using the triangle area formula these ratios can be shown to be equal to ). Since and are midpoints of and respectively, , so we are left with

By the angle bisector theorem, this ratio is equal to , where . We finally have

as required. It’s interesting that the answer has no dependence on the length of side .

Here is the original way I had solved the problem. Let be the origin and define vectors and of length and respectively. Then as is the angle bisector, we can write , where and .

Also let where . We find by using the fact that :

From this,

Since is on , for some between 0 and 1,

Since is also on the median , it and have equal components of and (being a median).

Hence, from which

We wish to find the ratio , which is as above.

Here is one easy way to find that I found only afterwards. If is the midpoint of then since is a right-angled triangle, and so , from which .

Then and are similar, from which

Hence and so

The original way I had found was without the benefit of this construction and by pure computation instead. Referring back to (1), we have and by the cosine rule, . Recalling that and ,

Secondly the denominator of (1) is

Combining (2) and (3) in (1) leads to the following feast of cancellation:

as was found before. Note that expressions (2) and (3) are respectively the square of the length of the angle bisector and the dot product in terms of the side lengths of any triangle, and we have found them to be similar in form.

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