Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 9"
(→‎Solution)
Line 2: Line 2:
 
<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>.
  
−
==Solution==
+
==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.
−
+
 
 +
==Solution 2 (Mass points)==
 +
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.
 +
 
 +
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.$
 +
 
 
==See Also==
 
==See Also==
 
{{Mock AIME box|year=Pre 2005|n=3|num-b=8|num-a=10}}
 
{{Mock AIME box|year=Pre 2005|n=3|num-b=8|num-a=10}}

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$ (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$CD=\frac{8.5}{8.5+6.5}\cdot 16.$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_3_Pre_2005_Problems/Problem_9&oldid=113544"