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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 9"

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==Solution 1==
 
==Solution 1==
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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 <math>\triangle</math>AND and <math>\triangle</math>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.
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Let the foot of the altitude from <math>C</math> to <math>AB</math> be point <math>F</math> and from <math>D</math> to <math>AB</math> be point <math>G.</math> <math>\triangle BAC \sim \triangle BAE.</math> Since <math>\triangle ABC</math> is isosceles, <math>\frac{BE}{AB} = \frac{1}{2},</math> so <math>BX = \frac{17}{2}.</math> Then <math>DX = 17 - 2 - \frac{17}{2} = \frac{13}{2}.</math> Notice that <math>AG</math> corresponds to <math>2</math>, <math>GF</math> corresponds to <math>\frac{13}{2},</math> and <math>BF</math> corresponds to <math>BX</math> by some scale factor <math>k</math>. Also note that <math>\triangle AGD \sim \triangle AFC.</math> Then we have <math>\frac{AG}{AF} = \frac{AD}{AC}</math> <math>\implies \frac{2k}{AB - BF} = \frac{AD}{16}</math> <math>\implies \frac{2k}{17k/2} = \frac{AD}{16}</math> <math>\implies \frac{4}{17} = \frac{AD}{16} \implies AD = \frac{64}{17}.</math> Then <math>CD</math> is <math>16 - \frac{64}{17} = \frac{208}{17}.</math> <math>208 + 17 = \boxed{225}.</math>
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~[[User:grogg007|grogg007]]
  
 
==Solution 2 (fakesolve, don't do this)==
 
==Solution 2 (fakesolve, don't do this)==

Latest revision as of 15:17, 10 July 2025

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 the foot of the altitude from $C$ to $AB$ be point $F$ and from $D$ to $AB$ be point $G.$ $\triangle BAC \sim \triangle BAE.$ Since $\triangle ABC$ is isosceles, $\frac{BE}{AB} = \frac{1}{2},$ so $BX = \frac{17}{2}.$ Then $DX = 17 - 2 - \frac{17}{2} = \frac{13}{2}.$ Notice that $AG$ corresponds to $2$, $GF$ corresponds to $\frac{13}{2},$ and $BF$ corresponds to $BX$ by some scale factor $k$. Also note that $\triangle AGD \sim \triangle AFC.$ Then we have $\frac{AG}{AF} = \frac{AD}{AC}$ $\implies \frac{2k}{AB - BF} = \frac{AD}{16}$ $\implies \frac{2k}{17k/2} = \frac{AD}{16}$ $\implies \frac{4}{17} = \frac{AD}{16} \implies AD = \frac{64}{17}.$ Then $CD$ is $16 - \frac{64}{17} = \frac{208}{17}.$ $208 + 17 = \boxed{225}.$

~grogg007

Solution 2 (fakesolve, don't do this)

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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