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 "2017 AMC 12B Problems/Problem 24"

(→‎See Also)
(→‎See Also)
Line 10: Line 10:
 
Let <math>CD=1</math>, <math>BC=x</math>, <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. The Pythagorean theorem states that <math>BD=sqrt(x^2+1)</math>. Since <math>BCD~ABC~CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=x^2/sqrt(x^2+1)</math> and <math>BE=x/sqrt(x^2+1)</math>. Let Point F be a point on <math>BC</math> such that <math>EF</math> is an altitude of triangle <math>CEB</math>. Note that <math>CEB~CFE~EFB</math>, so <math>BF</math> and <math>CF</math> can be calculated. Solving for these lengths gives <math>BF=x/(x^2+1)</math> and <math>CF=x^3/(x^2+1)</math>.
 
Let <math>CD=1</math>, <math>BC=x</math>, <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. The Pythagorean theorem states that <math>BD=sqrt(x^2+1)</math>. Since <math>BCD~ABC~CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=x^2/sqrt(x^2+1)</math> and <math>BE=x/sqrt(x^2+1)</math>. Let Point F be a point on <math>BC</math> such that <math>EF</math> is an altitude of triangle <math>CEB</math>. Note that <math>CEB~CFE~EFB</math>, so <math>BF</math> and <math>CF</math> can be calculated. Solving for these lengths gives <math>BF=x/(x^2+1)</math> and <math>CF=x^3/(x^2+1)</math>.
 
==See Also==
 
==See Also==
−
{{AMC12 box|year=2017|ab=B|num-b=23|after=25}}
+
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:26, 16 February 2017

Problem

Quadrilateral $ABCD$ has right angles at $B$ and $C$, Triangle $ABC$ ~ Triangle $BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that Triangle $ABC$ ~ Triangle $CEB$ and the area of Triangle $AED$ is $17$ times the area of Triangle $CEB$. What is $AB/BC$ $\textbf{(A) } 1+sqrt(2) \qquad \textbf{(B) } 2 + sqrt(2) \qquad \textbf{(C) } sqrt(17) \qquad \textbf{(D) } 2 + sqrt(5) \qquad \textbf{(E) } 1 + 2sqrt(3)$

Solution

Solution by TorrTar

Let $CD=1$, $BC=x$, $AB=x^2$. Note that $AB/BC=x$. The Pythagorean theorem states that $BD=sqrt(x^2+1)$. Since $BCD~ABC~CEB$, the ratios of side lengths must be equal. Since $BC=x$, $CE=x^2/sqrt(x^2+1)$ and $BE=x/sqrt(x^2+1)$. Let Point F be a point on $BC$ such that $EF$ is an altitude of triangle $CEB$. Note that $CEB~CFE~EFB$, so $BF$ and $CF$ can be calculated. Solving for these lengths gives $BF=x/(x^2+1)$ and $CF=x^3/(x^2+1)$.

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_24&oldid=83927"