Mock AIME 1 2006-2007 Problems/Problem 1

$\triangle ABC$ has positive integer side lengths of $x$ , $y$ , and $17$ . The angle bisector of $\angle BAC$ hits $BC$ at $D$ . If $\angle C=90^\circ$ , and the maximum value of $\frac{[ABD]}{[ACD]}=\frac{m}{n}$ where $m$ and $n$ are relatively prime positive intgers, find $m+n$ . (Note $[ABC]$ denotes the area of $\triangle ABC$ ).

Solution

Assume without loss of generality that $x \leq y$ . Then the hypotenuse of right triangle $\triangle ABC$ either has length 17, in which case $x^2 + y^2 = 17$ , or has length $y$ , in which case $x^2 + 17^2 = y^2$ , by the Pythagorean Theorem.

In the first case, you can either know your Pythagorean triples or do a bit of casework to find that the only solution is $x = 8, y = 15$ . In the second case, we have $17^2 = y^2 - x^2 = (y - x)(y + x)$ , a factorization as a product of two different positive integers, so we must have $y - x = 1$ and $y + x = 17^2 = 289$ from which we get the solution $x = 144, y= 145$ .

Now, note that the area $[ABD] = \frac 12 AB \cdot AD \cdot \sin BAD$ and $[ACD] = \frac 12 \cdot AD \cdot AC \cdot \sin CAD$ , and since $AD$ is an angle bisector we have $\displaystyle \angle BAD = \angle CAD$ so $\frac{[ABD]}{[ACD]} = \frac{AB}{AC}$ .

In our first case, this value may be either $\frac {17}{8}$ or $\frac{17}{15}$ . In the second, it may be either $\frac{145}{144}$ or $\frac{145}{17}$ . Of these four values, the last is clearly the greatest. 17 and 145 are relatively prime, so our answer is $17 + 145 = 162$ .

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Mock AIME 1 2006-2007

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Mock AIME 1 2006-2007 Problems/Problem 1

Solution