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

2019 AIME II Problems/Problem 11

Revision as of 08:40, 17 June 2022 by Vvsss (talk | contribs) (→‎Solution 6 (Inversion simplified))

Problem

Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

[asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("$A$",A,dir(105)); label("$B$",B,dir(-135)); label("$C$",C,dir(-75)); dot((2.68,2.25)); label("$K$",(2.68,2.25),dir(-150)); label("$\omega_1$",(-6,1)); label("$\omega_2$",(14,6)); label("$7$",(A+B)/2,dir(140)); label("$8$",(B+C)/2,dir(-90)); label("$9$",(A+C)/2,dir(60)); [/asy] -Diagram by Brendanb4321


Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$, so $\triangle AKB\sim \triangle CKA$. Using similarity ratios, we can easily find \[AK^2=BK*KC\] However, since $AB=7$ and $CA=9$, we can use similarity ratios to get \[BK=\frac{7}{9}AK, CK=\frac{9}{7}AK\]

  • Now we use Law of Cosines on $\triangle AKB$: From reverse Law of Cosines, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}$. This gives us \[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\] \[\implies \frac{196}{81}AK^2=49\] \[AK=\frac{9}{2}\] so our answer is $9+2=\boxed{011}$.

-franchester

  • The motivation for using the Law of Cosines ("LoC") is after finding the similar triangles it's hard to figure out what to do with $BK$ and $CK$ yet we know $BC$ which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealing with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.

-First

11111111:L)xiexie

Solution 2 (Inversion)

Consider an inversion with center $A$ and radius $r=AK$. Then, we have $AB\cdot AB^*=AK^2$, or $AB^*=\frac{AK^2}{7}$. Similarly, $AC^*=\frac{AK^2}{9}$. Notice that $AB^*KC^*$ is a parallelogram, since $\omega_1$ and $\omega_2$ are tangent to $AC$ and $AB$, respectively. Thus, $AC^*=B^*K$. Now, we get that \[\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}\] so by Law of Cosines on $\triangle AB^*K$ we have \[(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)\] \[\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}\] \[\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{22AK^2}{63\cdot21}\] \[\Rightarrow AK=\frac{9}{2}\] Then, our answer is $9+2=\boxed{11}$. -brianzjk

Solution 3 (Death By Trig Bash)

14. Let the centers of the circles be $O_{1}$ and $O_{2}$ where the $O_{1}$ has the side length $7$ contained in the circle. Now let $\angle BAC =x.$ This implies \[\angle AO_{1}B = \angle AO_{2}C = 2x\] by the angle by by tangent. Then we also know that \[\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x\] Now we first find $\cos x.$ We use law of cosines on $\bigtriangleup ABC$ to obtain \[64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}\] \[\implies \cos{x} =\frac{11}{21}\] \[\implies \sin{x} =\frac{8\sqrt{5}}{21}\] Then applying law of sines on $\bigtriangleup AO_{1}B$ we obtain \[\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}}\] \[\implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}}\] \[\implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}\] Using similar logic we obtain $OA_{1} =\frac{189}{16\sqrt{5}}.$

Now we know that $\angle O_{1}AO_{2}=180^{\circ}-x.$ Thus using law of cosines on $\bigtriangleup O_{1}AO_{2}$ yields \[O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}\] While this does look daunting we can write the above expression as \[\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}\] Then factoring yields \[\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}\] The area \[[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] Now $AK$ is twice the length of the altitude of $\bigtriangleup O_{1}AO_{2}$ so we let the altitude be $h$ and we have \[\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] \[\implies h =\frac{9}{4}\] Thus our desired length is $\frac{9}{2} \implies n+n = \boxed{11}.$

Solution 4 (Video)

Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI

Solution 5 (Olympiad Geometry)

By the definition of $K$, it is the spiral center mapping $BA\to AC$, which means that it is the midpoint of the $A$-symmedian chord. In particular, if $M$ is the midpoint of $BC$ and $M'$ is the reflection of $A$ across $K$, we have $\triangle ABM'\sim\triangle AMC$. By Stewart's Theorem, it then follows that \[AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.\]

Solution 6 (Inversion simplified)

2019 AIME II 11.png

The median of $\triangle ABC$ is $AM = \sqrt{\frac {AB^2 + AC^2 }{2} – \frac{BC^2}{4}} = 7.$

Consider an inversion with center $A$ and radius $AK$ (inversion with respect the red circle). Let $K, B',$ and $C'$ be inverse points for $K, B,$ and $C,$ respectively.

Image of line $AB$ is line $AB, B'$ lies on this line.

Image of $\omega_2$ is line $KC'||AB$ (circle $\omega_2$ passes through K, C and is tangent to the line $AB$ at point $A.$ Diagram shows cicle and it’s image using same color).

Similarly, $AC||B'K (B'K$ is the image of the circle $\omega_1$).

Therefore $AB'KC'$ is a parallelogram, $AF$ is median of $\triangle AB'C'$ and $AK = 2 AF.$ Then, we have $AB'=\frac{AK^2}{7}$. $\triangle ABC \sim \triangle AC'B'$ with coefficient $k =\frac {AB'}{AC} = \frac{AK^2}{7\cdot 9}.$

So median \[AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot \frac{AK^2}{7\cdot 9} \implies AK = \frac{9}{2}.\] Shelomovskii, vvsss, www.deoma-cmd.ru

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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=2019_AIME_II_Problems/Problem_11&oldid=174979"