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 "2001 AIME II Problems/Problem 7"

(→‎Solution 4 (easy but hard to see))
 
Line 48: Line 48:
 
The problem can be reduced to a <math>3-4-5</math> triangle for the initial calculations, where <math>C_2</math> is calculated to be (<math>\frac{5}{2}, \frac{1}{2})</math>, and <math>C_3</math> is calculated to be (<math>\frac{1}{3}, \frac{7}{3})</math>. After we find the incenters the points can be scaled up by a factor of <math>30</math> for the final distance calculation.
 
The problem can be reduced to a <math>3-4-5</math> triangle for the initial calculations, where <math>C_2</math> is calculated to be (<math>\frac{5}{2}, \frac{1}{2})</math>, and <math>C_3</math> is calculated to be (<math>\frac{1}{3}, \frac{7}{3})</math>. After we find the incenters the points can be scaled up by a factor of <math>30</math> for the final distance calculation.
  
−
== Solution 4 (easy but hard to see) ==  
+
=== Solution 4 (easy but hard to see) ===
 
We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon <math>PSTVU</math> is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with <math>C_3C_2</math> as its hypotenuse. The right angle will be at point <math>Y</math>. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of <math>C_3Y</math> is <math>50 + 15 = 65</math>, as seen by the inradius of <math>C_2</math> and <math>10</math> less than the square's side length. <math>C_2Y</math> is <math>45 + 10 = 55</math>, which is <math>15</math> less than the square plus the inradius of <math>C_3</math>. Our final answer is <math>\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}</math>.
 
We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon <math>PSTVU</math> is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with <math>C_3C_2</math> as its hypotenuse. The right angle will be at point <math>Y</math>. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of <math>C_3Y</math> is <math>50 + 15 = 65</math>, as seen by the inradius of <math>C_2</math> and <math>10</math> less than the square's side length. <math>C_2Y</math> is <math>45 + 10 = 55</math>, which is <math>15</math> less than the square plus the inradius of <math>C_3</math>. Our final answer is <math>\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}</math>.
  

Latest revision as of 01:15, 26 December 2022

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, the inradius could be 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. (This formula should be used only for right triangles.) 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 $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$.


Solution 3

The radius of an incircle is $r=A_t/\text{semiperimeter}$. The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$. The radius, therefore, is equal to $\frac{5400}{180} = 30$. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$, $\frac{1}{2}(90) = 45$, and $\frac{1}{2}\times150 = 75$. The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$.


Defining $P$ as $(0,0)$, $C_2$ is equal to $(60+15,15)$ or $(75,15)$. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$, $\frac{1}{3}\times120$, $\frac{1}{3}\times150$ or $30,40,50$. The radius of $C_3$ by using the formula mentioned at the beginning is $10$. Using $P$ as $(0,0)$, $C_3$ is equal to $(10, 60+10)$ or $(10,70)$. Using the distance formula, the distance between $C_2$ and $C_3$: $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$, thus $n$ is $\boxed{725}$.

Note

The problem can be reduced to a $3-4-5$ triangle for the initial calculations, where $C_2$ is calculated to be ($\frac{5}{2}, \frac{1}{2})$, and $C_3$ is calculated to be ($\frac{1}{3}, \frac{7}{3})$. After we find the incenters the points can be scaled up by a factor of $30$ for the final distance calculation.

Solution 4 (easy but hard to see)

We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$, as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$, which is $15$ less than the square plus the inradius of $C_3$. Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$.

See also

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

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=2001_AIME_II_Problems/Problem_7&oldid=184868"