2001 AIME I Problems/Problem 7

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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

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,...

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 (mass points)

pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4");pair B=(0,0), C=(20,0), ...

Let $P$ be the incircle; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$ , and name the 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, $m+n=\boxed{923}$ .

2001 AIME I (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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