2001 AIME II Problems/Problem 7

Problem

Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ , $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$ ?

Solution

Solution 1 (analytic)

$[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); [/asy]$

Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$ , where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$ ), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$ . Or, you could find the inradius directly by using the formula $\frac{a+b-c}{2}$ , where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (Can you see why? Plus, use this formula only for right triangles because these are actually the only cases.) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$ , and so $RS = 60, UQ = 30$ .

Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$ . Since the ratio of corresponding lengths of similar figures are the same, we have

$\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.$

Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$ , respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$ . Therefore, the answer is $n = \boxed{725}$ .

Solution 2 (synthetic)

$[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); pair A2 = IP(incircle(R,S,T), Q--R), A3 = IP(incircle(Q,U,V), Q--R); D(D(MP("A_2",A2,NE)) -- O2, linetype("4 4")+linewidth(0.6)); D(D(MP("A_3",A3,NE)) -- O3 -- foot(O3, A2, O2), linetype("4 4")+linewidth(0.6)); [/asy]$

We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$ .

By the Two Tangent Theorem, we find that $A_{1}Q = 60$ , $A_{1}R = 90$ . Using the similar triangles, $RA_{2} = 45$ , $QA_{3} = 20$ , so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$ . Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$ .

2001 AIME II (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

Page

Toolbox

Search

2001 AIME II Problems/Problem 7

Problem

Contents

Solution

Solution 1 (analytic)

Solution 2 (synthetic)

See also