Mock AIME 3 Pre 2005 Problems/Problem 9

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$ABC$ is an isosceles triangle with base $\overline{AB}$. $D$ is a point on $\overline{AC}$ and $E$ is the point on the extension of $\overline{BD}$ past $D$ such that $\angle{BAE}$ is right. If $BD = 15, DE = 2,$ and $BC = 16$, then $CD$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $m + n$.

Solution 1

Let AB=x. Call the foot of the perpendicular from D to AB N, and the foot of the perpendicular from C to AB M. By similarity, AN=2x/17. Also, AM=x/2. Since $\triangle$AND and $\triangle$CAM are similar, we have (2x/17)/AD=(x/2)/16. Hence, AD=64/17, and CD=16-AD=208/17, so the answer is 225.

Solution 2 (Mass points)

Let the perpendicular from $C$ intersect $AB$ at $H.$ Let $CH$ intersect $BD$ at $P.$ Then let $AP$ intersect $BC$ at $F.$

Note that $\triangle AEB\sim \triangle HPB,$ with a factor of $2.$ So $BP=8.5$ and $DP=6.5.$ Then the mass of $P$ is $15$ and the mass of $D$ is $8.5$ and the mass of $B$ is $6.5.$ Because the triangle is isosceles, the mass of $A$ is also $6.5.$ So $CD=\frac{8.5}{8.5+6.5}\cdot 16.$

See Also

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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