Mock AIME 3 Pre 2005 Problems/Problem 9

Problem

$ABC$ 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$ (Error compiling LaTeX. Unknown error_msg)\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.$

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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Mock AIME 3 Pre 2005 Problems/Problem 9

Contents

Problem

Solution 1

Solution 2 (Mass points)

See Also