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 "2002 AMC 12A Problems/Problem 22"

(→‎Solution)
(→‎Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 29: Line 29:
 
==Solution==
 
==Solution==
  
−
Clearly <math>BC=5</math> and <math>AC=5\sqrt{3}</math>. Choose a <math>P'</math> and get a corresponding <math>D'</math> such that <math>BD'= 5\sqrt{2}</math> and <math>AD'=5</math>. For <math> BD > 5\sqrt2 </math> we need <math>CD>5</math>, creating an isoclese right triangle with hyptonuse <math>5\sqrt {2}</math> . Thus the point <math>P</math> may only lie in the triangle <math>ABD'</math>. The probability of it doing so is the ratio of areas of <math>ABD'</math> to <math>ABC</math>, or equivalently, the ratio of <math>AD'</math> to <math>AC</math> because the triangles have identical altitudes when taking <math>AD'</math> and <math>AC</math> as bases. This ratio is equal to <math>\frac{AC-BD'}{AC}=1-\frac{BD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}</math>. Thus the answer is <math>\boxed{C}</math>.
+
Clearly <math>BC=5</math> and <math>AC=5\sqrt{3}</math>. Choose a <math>P'</math> and get a corresponding <math>D'</math> such that <math>BD'= 5\sqrt{2}</math> and <math>CD'=5</math>. For <math> BD > 5\sqrt2 </math> we need <math>CD>5</math>, creating an isosceles right triangle with hypotenuse <math>5\sqrt {2}</math> . Thus the point <math>P</math> may only lie in the triangle <math>ABD'</math>. The probability of it doing so is the ratio of areas of <math>ABD'</math> to <math>ABC</math>, or equivalently, the ratio of <math>AD'</math> to <math>AC</math> because the triangles have identical altitudes when taking <math>AD'</math> and <math>AC</math> as bases. This ratio is equal to <math>\frac{AC-CD'}{AC}=1-\frac{CD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}</math>. Thus the answer is <math>\boxed{C}</math>.
 +
 
 +
== Video Solution ==
 +
https://youtu.be/WH38DzdClKM
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2002|ab=A|num-b=21|num-a=23}}
 
{{AMC12 box|year=2002|ab=A|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 10:11, 18 July 2023

Problem

Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$ , and $AB = 10$. Let $P$ be randomly chosen inside $ABC$ , and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. What is the probability that $BD > 5\sqrt2$?

[asy] import math; unitsize(4mm); defaultpen(fontsize(8pt)+linewidth(0.7)); dotfactor=4; pair A=(10,0); pair C=(0,0); pair B=(0,10.0/sqrt(3)); pair P=(2,2); pair D=extension(A,C,B,P); draw(A--C--B--cycle); draw(B--D); dot(P); label("A",A,S); label("D",D,S); label("C",C,S); label("P",P,NE); label("B",B,N);[/asy]


$\textbf{(A)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{3-\sqrt3}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{5-\sqrt5}{5}$

Solution

Clearly $BC=5$ and $AC=5\sqrt{3}$. Choose a $P'$ and get a corresponding $D'$ such that $BD'= 5\sqrt{2}$ and $CD'=5$. For $BD > 5\sqrt2$ we need $CD>5$, creating an isosceles right triangle with hypotenuse $5\sqrt {2}$ . Thus the point $P$ may only lie in the triangle $ABD'$. The probability of it doing so is the ratio of areas of $ABD'$ to $ABC$, or equivalently, the ratio of $AD'$ to $AC$ because the triangles have identical altitudes when taking $AD'$ and $AC$ as bases. This ratio is equal to $\frac{AC-CD'}{AC}=1-\frac{CD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}$. Thus the answer is $\boxed{C}$.

Video Solution

https://youtu.be/WH38DzdClKM

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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=2002_AMC_12A_Problems/Problem_22&oldid=195862"