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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 9"
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<math>ABC</math> is an isosceles triangle with base <math>\overline{AB}</math>. <math>D</math> is a point on <math>\overline{AC}</math> and <math>E</math> is the point on the extension of <math>\overline{BD}</math> past <math>D</math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Determine <math>m + n</math>.
 
<math>ABC</math> is an isosceles triangle with base <math>\overline{AB}</math>. <math>D</math> is a point on <math>\overline{AC}</math> and <math>E</math> is the point on the extension of <math>\overline{BD}</math> past <math>D</math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Determine <math>m + n</math>.
  
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==Solution==
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==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 <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.
 
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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==See also==
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==Solution 2 (Mass points)==
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Let the perpendicular from <math>C</math> intersect <math>AB</math> at <math>H.</math> Let <math>CH</math> intersect <math>BD</math> at <math>P.</math> Then let <math>AP</math> intersect <math>BC</math> at <math>F.</math>
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Note that <math>\triangle AEB\sim \triangle HPB,</math> with a factor of <math>2.</math> So <math>BP=8.5</math> and <math>DP=6.5.</math> Then the mass of <math>P</math> is <math>15</math> and the mass of <math>D</math> is <math>8.5</math> and the mass of <math>B</math> is <math>6.5.</math> Because the triangle is isosceles, the mass of <math>A</math> is also <math>6.5.</math> So <math>CD=\frac{8.5}{8.5+6.5}\cdot 16.</math>
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==See Also==
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{{Mock AIME box|year=Pre 2005|n=3|num-b=8|num-a=10}}

Latest revision as of 15:32, 27 December 2019

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 $\triangle AEB\sim \triangle HPB,$ 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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