2019 AIME II Problems/Problem 11


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),2*down); label("$\omega_1$",(-4.5,1)); label("$\omega_2$",(12.75,6)); 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, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=\angle AKB=-\frac{11}{21}$

Giving 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 $9+2=\boxed{011}$.

$^{(*)}$ Let $O$ be the center of $\omega_1$. Then $\angle KAC = 90 - \angle OAK = 90 - \frac{1}{2}(180 - \angle AOK) = \frac{\angle AOK}{2} = \angle ABK$. Thus, $\angle ABK = \angle KAC$

-franchester; $^{(*)}$ by firebolt360


  • In order to get to the Law of Cosines first, we first apply the LOC to $\triangle{ABC},$ obtaining $\angle{BAC}.$
  • We angle chase before applying the law of cosines to $\angle{AKB}.$

Note that $\angle{ABK}=\angle{KAC}$ and $\angle{KCA}=\angle{KAB}$ from tangent-chord.

Thus, $\angle{AKC}=\angle{AKB}=180^{\circ}-(\angle{ABK}+\angle{KAB}).$

However from our tangent chord, note that: \[\angle{ABK}+\angle{KAB}=\angle{KAC}+\angle{KAB}=\angle{BAC}.\] Thus, $\angle{AKB}=180^\circ-\angle{BAC}.$

  • As an alternative approach, note that the sum of the angles in quadrilateral $ABKC$ is $360^{\circ}$ and we can find $\angle{AKB}=\frac12$ of convex $\angle{BKC},$ which is just:

\[\frac12 \left(360^{\circ}-2(\angle{KAB}+\angle{KBA}\right) = 180^\circ - \angle{BAC}.\]


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)

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 \[\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x\] by the angle by by tangent. Then we also know that \[\angle AO_{1}B = \angle AO_{2}C = 2x\] 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 \[\frac{7}{\sin{2x}} =\frac{O_{1}B}{\sin{90^{\circ}-x}}\] \[\implies\frac{7}{2\sin{x}\cos{x}} =\frac{O_{1}B}{\cos{x}}\] \[\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}\] Using similar logic we obtain $O_{2}A =\frac{189}{16\sqrt{5}}.$

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 \[\frac{1}{2} \cdot h \cdot\frac{147}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}\] \[\implies h =\frac{9}{4}\] Thus our desired length is $\frac{9}{2} \implies m+n = \boxed{11}.$

-minor edits by faliure167

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)


The median of $\triangle ABC$ is $AM = \sqrt{\frac {AB^2 +  AC^2 }{2} – \frac{BC^2}{4}} = 7$ (via Stewart's Theorem).

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 circle and its 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 \[AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot  \frac{AK^2}{7\cdot 9} \implies  AK = \frac{9}{2}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution 7 (Heavy Bash)

We start by assigning coordinates to point $A$, labeling it $(0,0)$ and point $B$ at $(7,0)$, and letting point $C$ be above the $x$-axis. Through an application of the Pythagorean Theorem and dropping an altitude to side $AB$, it is easy to see that $C$ has coordinates $(33/7, 24\sqrt{5}/7)$.

Let $O1$ be the center of circle $\omega_1$ and $O2$ be the center of circle $\omega_2$. Since circle $\omega_1$ contains both points $A$ and $B$, $O1$ must lie on the perpendicular bisector of line $AB$, and similarly $O2$ must lie on the perpendicular bisector of line $AC$. Through some calculations, we find that the perpendicular bisector of $AB$ has equation $x = 3.5$, and the perpendicular bisector of $AC$ has equation $y = {-11\sqrt{5}/40 \cdot x} + 189\sqrt{5}/80$.

Since circle $\omega_1$ is tangent to line $AC$ at $A$, its radius must be perpendicular to $AC$ at $A$. Therefore, the radius has equation $y = {{-11\cdot\sqrt{5}/40} \cdot x}$. Substituting the $x$-coordinate of $O1$ into this, we find the y-coordinate of $O1 = {{-11 \cdot \sqrt{5}/40} \cdot 7/2} = {-77 \cdot \sqrt{5}/80}$.

Similarly, since circle $\omega_2$ is tangent to line $AB$ at $A$, its radius must be perpendicular to $AB$ at $A$. Therefore, the radius has equation $x = 0$ and combining with the previous result for $O2$ we get that the coordinates of $O2$ are $(0, 189\sqrt{5}/80)$.

We now find the slope of $O1O2$, the line joining the centers of circles $\omega_1$ and $\omega_2$, which turns out to be ${({266 \cdot \sqrt{5} / 80}) \cdot -2/7} = {-19 \cdot \sqrt{5}/20}$. Since the $y$-intercept of that line is at $O2(0,189\sqrt{5}/80)$, the equation is $y = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}$. Since circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $K$, line $AK$ is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles, $AK$ has slope ${4 \cdot \sqrt{5}/19}$. Since point $A$ is $(0,0)$, this line has a $y$-intercept of $0$, so it has equation $y$ = ${{4 \cdot \sqrt{5}/19} \cdot x}$.

We set ${{4 \cdot \sqrt{5}/19} \cdot x} = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}$ in order to find the intersection $I$ of the radical axis $AK$ and $O1O2$. Through some moderate bashing, we find that the intersection point is $I(57/28, {3 \cdot \sqrt{5}/7})$. We know that either intersection point of two circles is the same distance from the intersection of radical axis and line joining the centers of those circles, so reflecting $A$ over $I$ yields $K$ and $AK$ = $2AI$ = (This is the most tedious part of the bash) ${2 \cdot \sqrt{(57/28)^2 + ({3 \cdot \sqrt{5}/7)^2)}}} = {2 \cdot \sqrt{3969/784}} = {2 \cdot 63/28} = {2 \cdot 9/4} = 9/2$. Therefore the answer is $9 + 2 = \boxed{011}.$

See Also

2019 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions

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