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

2001 AIME I Problems/Problem 7

Revision as of 21:03, 25 July 2008 by Newton1988 (talk | contribs) (→‎Solution)

Problem

Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Solution 1

[asy] pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW)); D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); [/asy]

The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$. By Heron's formula, the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}2$. Using the formula $A = rs$, we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$. Since $\triangle ADE \sim \triangle ABC$, the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$. From $A=\frac{1}{2}bh$, we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$. Thus, we have

$\frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.$

Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$.

Solution 2

Since the point is the center (call it P) of the inscribed circle, it must be the intersection of all three angle bisectors. Drawing the bisector AP, to where it intersects BC, we shall call this intersection F. Using the angle bisector theorem, we know the ratio BF:CF is 21:22, thus we shall assign a weight of 22 to point B and a weight of 21 to point C, giving F a weight of 43. In the same manner, using another bisector, we find that A has a weight of 20. So, now we know P has a weight of 63, and the ratio of FP:PA is 20:43. Therefore, the smaller similar triangle ADE is 43/63 the height of the original triangle ABC. So, DE is 43/63 the size of BC. Multiplying this ratio by the length of BC, we find DE is 860/63 = m/n. Therefore, m+n=923.

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_7&oldid=27191"