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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 1"
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In the first case, you can either know your [[Pythagorean triple]]s or do a bit of casework to find that the only solution is <math>x = 8, y = 15</math>.  In the second case, we have <math>17^2 = y^2 - x^2 = (y - x)(y + x)</math>, a [[factor]]ization as a product of two different [[positive integer]]s, so we must have <math>y - x = 1</math> and <math>y + x = 17^2 = 289</math> from which we get the solution <math>x = 144, y= 145</math>.
 
In the first case, you can either know your [[Pythagorean triple]]s or do a bit of casework to find that the only solution is <math>x = 8, y = 15</math>.  In the second case, we have <math>17^2 = y^2 - x^2 = (y - x)(y + x)</math>, a [[factor]]ization as a product of two different [[positive integer]]s, so we must have <math>y - x = 1</math> and <math>y + x = 17^2 = 289</math> from which we get the solution <math>x = 144, y= 145</math>.
  
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Now, note that the [[area]] <math>[ABD] = \frac 12 AB \cdot AD \cdot \sin BAD</math> and <math>[ACD] = \frac 12 \cdot AD \cdot AC \cdot \sim CAD</math>, and since <math>AD</math> is an [[angle bisector]] we have <math>\angle BAD = \angle CAD</math> so <math>\frac{[ABD]}{[ACD]} = \frac{AB}{AC}</math>.
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Now, note that the [[area]] <math>[ABD] = \frac 12 AB \cdot AD \cdot \sin BAD</math> and <math>[ACD] = \frac 12 \cdot AD \cdot AC \cdot \sim CAD</math>, and since <math>AD</math> is an [[angle bisector]] we have <math>\displaystyle \angle BAD = \angle CAD</math> so <math>\frac{[ABD]}{[ACD]} = \frac{AB}{AC}</math>.
  
 
In our first case, this value may be either <math>\frac {17}{8}</math> or <math>\frac{17}{15}</math>.  In the second, it may be either <math>\frac{145}{144}</math> or <math>\frac{145}{17}</math>.  Of these four values, the last is clearly the greatest.  17 and 145 are [[relatively prime]], so our answer is <math>17 + 145 = 162</math>.
 
In our first case, this value may be either <math>\frac {17}{8}</math> or <math>\frac{17}{15}</math>.  In the second, it may be either <math>\frac{145}{144}</math> or <math>\frac{145}{17}</math>.  Of these four values, the last is clearly the greatest.  17 and 145 are [[relatively prime]], so our answer is <math>17 + 145 = 162</math>.

Revision as of 15:05, 23 August 2006

$\triangle ABC$ 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 $[ACD] = \frac 12 \cdot AD \cdot AC \cdot \sim CAD$, 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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