Triangle ABC has positive integer side lengths. One of the side is 17 units in length. The angle bisector of angle BAC intersects BC at D. If angle C equals 90 degrees, find the maximum value of (area of ABD)/(area of ACD).
ABC is a right triangle with a side of length 17.
The ratio of the areas is equal to the ratio of their bases BD/DC, since they have the same height AC.
The ABC can only be an 8-15-17.
Therefore AB=17 and the other sides give two cases:
Case 1 AC=15, BC=8.
CD=15tan(0.5arctan(8/15)) = 15/4
BD=17/4
The ratio is 17/15
Case 2 AC=8, BC=15.
CD=15tan(0.5arctan(15/8)) = 24/5
BD=51/5
The ratio is 17/8.
This is the greater of the two and thus the solution.
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Posted by Jer
on 2024-11-17 12:36:53 |
Solution
