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 "2019 AIME II Problems/Problem 11"

m (→‎Solution 2)
(→‎Solution 3 (Death By Trig Bash))
 
(49 intermediate revisions by 16 users not shown)
Line 15: Line 15:
 
label("$C$",C,dir(-75));
 
label("$C$",C,dir(-75));
 
dot((2.68,2.25));
 
dot((2.68,2.25));
−
label("$K$",(2.68,2.25),dir(-150));
+
label("$K$",(2.68,2.25),2*down);
−
label("$\omega_1$",(-6,1));
+
label("$\omega_1$",(-4.5,1));
−
label("$\omega_2$",(14,6));
+
label("$\omega_2$",(12.75,6));
 
label("$7$",(A+B)/2,dir(140));
 
label("$7$",(A+B)/2,dir(140));
 
label("$8$",(B+C)/2,dir(-90));
 
label("$8$",(B+C)/2,dir(-90));
Line 25: Line 25:
  
  
−
Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math>. This gives us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>.
+
Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math><math>^{(*)}</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath>  
−
-franchester
 
  
−
==Solution 2==
+
*Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=\angle AKB=-\frac{11}{21}</math>
 +
Giving us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>.
  
−
On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=nJydO5CLuuI
+
<math>^{(*)}</math> Let <math>O</math> be the center of <math>\omega_1</math>. Then <math>\angle KAC = 90 - \angle OAK = 90 - \frac{1}{2}(180 - \angle AOK) = \frac{\angle AOK}{2} = \angle ABK</math>. Thus, <math>\angle ABK = \angle KAC</math>
  
−
Please like, share, and subscribe!
+
-franchester; <math>^{(*)}</math> by firebolt360
 +
===Supplement===
 +
*In order to get to the Law of Cosines first, we first apply the LOC to <math>\triangle{ABC},</math> obtaining <math>\angle{BAC}.</math>
 +
*We angle chase before applying the law of cosines to <math>\angle{AKB}.</math>
 +
 
 +
Note that <math>\angle{ABK}=\angle{KAC}</math> and <math>\angle{KCA}=\angle{KAB}</math> from tangent-chord.
 +
 
 +
Thus, <math>\angle{AKC}=\angle{AKB}=180^{\circ}-(\angle{ABK}+\angle{KAB}).</math>
 +
 
 +
However from our tangent chord, note that:
 +
<cmath>\angle{ABK}+\angle{KAB}=\angle{KAC}+\angle{KAB}=\angle{BAC}.</cmath>
 +
Thus, <math>\angle{AKB}=180^\circ-\angle{BAC}.</math>
 +
 
 +
*As an alternative approach, note that the sum of the angles in quadrilateral <math>ABKC</math> is <math>360^{\circ}</math> and we can find <math>\angle{AKB}=\frac12</math> of convex <math>\angle{BKC},</math> which is just:
 +
<cmath>\frac12 \left(360^{\circ}-2(\angle{KAB}+\angle{KBA}\right) = 180^\circ - \angle{BAC}.</cmath>
 +
 
 +
~mathboy282
 +
 
 +
==Solution 2 (Inversion)==
 +
Consider an inversion with center <math>A</math> and radius <math>r=AK</math>. Then, we have <math>AB\cdot AB^*=AK^2</math>, or <math>AB^*=\frac{AK^2}{7}</math>. Similarly, <math>AC^*=\frac{AK^2}{9}</math>. Notice that <math>AB^*KC^*</math> is a parallelogram, since <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AC</math> and <math>AB</math>, respectively. Thus, <math>AC^*=B^*K</math>. Now, we get that
 +
<cmath>\cos(\angle AB^*K)=\cos(180-\angle BAC)=-\frac{11}{21}</cmath>
 +
so by Law of Cosines on <math>\triangle AB^*K</math> we have
 +
<cmath>(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)</cmath>
 +
<cmath>\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}</cmath>
 +
<cmath>\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{22AK^2}{63\cdot21}</cmath>
 +
<cmath>\Rightarrow AK=\frac{9}{2}</cmath>
 +
Then, our answer is <math>9+2=\boxed{11}</math>.
 +
-brianzjk
 +
 
 +
== Solution 3 (Death By Trig Bash) ==
 +
 
 +
Let the centers of the circles be <math>O_{1}</math> and <math>O_{2}</math> where the <math>O_{1}</math> has the side length <math>7</math> contained in the circle. Now let <math>\angle BAC =x.</math> This implies <cmath>\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x</cmath> by the angle by by tangent. Then we also know that <cmath>\angle AO_{1}B = \angle AO_{2}C = 2x</cmath> Now we first find <math>\cos x.</math> We use law of cosines on <math>\bigtriangleup ABC</math> to obtain <cmath>64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}</cmath> <cmath>\implies \cos{x} =\frac{11}{21}</cmath> <cmath>\implies \sin{x} =\frac{8\sqrt{5}}{21}</cmath> Then applying law of sines on <math>\bigtriangleup AO_{1}B</math> we obtain <cmath>\frac{7}{\sin{2x}} =\frac{O_{1}B}{\sin{90^{\circ}-x}}</cmath> <cmath>\implies\frac{7}{2\sin{x}\cos{x}} =\frac{O_{1}B}{\cos{x}}</cmath> <cmath>\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>O_{2}A =\frac{189}{16\sqrt{5}}.</math>
 +
 
 +
Now we know that <math>\angle O_{1}AO_{2}=180^{\circ}-x.</math> Thus using law of cosines on <math>\bigtriangleup O_{1}AO_{2}</math> yields <cmath>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}}</cmath> While this does look daunting we can write the above expression as <cmath>\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)}</cmath> Then factoring yields <cmath>\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}</cmath> The area <cmath>[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}</cmath> Now <math>AK</math> is twice the length of the altitude of <math>\bigtriangleup O_{1}AO_{2}</math> so we let the altitude be <math>h</math> and we have <cmath>\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}</cmath> <cmath>\implies h =\frac{9}{4}</cmath> Thus our desired length is <math>\frac{9}{2} \implies m+n = \boxed{11}.</math>
 +
 
 +
-minor edits by faliure167
 +
 
 +
==Solution 4 (Video)==
 +
 
 +
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI
 +
 
 +
==Solution 5 (Olympiad Geometry)==
 +
 
 +
By the definition of <math>K</math>, it is the spiral center mapping <math>BA\to AC</math>, which means that it is the midpoint of the <math>A</math>-symmedian chord. In particular, if <math>M</math> is the midpoint of <math>BC</math> and <math>M'</math> is the reflection of <math>A</math> across <math>K</math>, we have <math>\triangle ABM'\sim\triangle AMC</math>. By Stewart's Theorem, it then follows that
 +
<cmath>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}.</cmath>
 +
 
 +
==Solution 6 (Inversion simplified)==
 +
[[File:AIME-II-2019-11.png|500px|right]]
 +
The median of <math>\triangle ABC</math> is <math>AM = \sqrt{\frac {AB^2 +  AC^2 }{2} – \frac{BC^2}{4}} = 7</math> (via Stewart's Theorem).
 +
 
 +
Consider an inversion with center <math>A</math> and radius <math>AK</math> (inversion with respect the red circle).
 +
Let <math>K, B',</math> and <math>C'</math> be inverse points for <math>K, B,</math> and <math>C,</math> respectively.
 +
 +
Image of line <math>AB</math> is line <math>AB, B'</math> lies on this line.
 +
 
 +
Image of <math>\omega_2</math> is line <math>KC'||AB</math> (circle <math>\omega_2</math> passes through K, C and is tangent to the line <math>AB</math> at point <math>A.</math> Diagram shows circle and its image using same color).
 +
 
 +
Similarly, <math>AC||B'K (B'K</math> is the image of the circle <math>\omega_1</math>).
 +
 +
Therefore <math>AB'KC'</math> is a parallelogram, <math>AF</math> is median of <math>\triangle AB'C'</math> and <math>AK = 2 AF.</math>
 +
Then, we have <math>AB'=\frac{AK^2}{7}</math>.  <math>\triangle ABC \sim \triangle AC'B'</math> with coefficient <math>k =\frac {AB'}{AC} = \frac{AK^2}{7\cdot 9}.</math>
 +
 
 +
So median <cmath>AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot  \frac{AK^2}{7\cdot 9} \implies  AK = \frac{9}{2}.</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
== Solution 7 (Heavy Bash) ==
 +
We start by assigning coordinates to point <math>A</math>, labeling it <math>(0,0)</math> and point <math>B</math> at <math>(7,0)</math>, and letting point <math>C</math> be above the <math>x</math>-axis. Through an application of the Pythagorean Theorem and dropping an altitude to side <math>AB</math>, it is easy to see that <math>C</math> has coordinates <math>(33/7, 24\sqrt{5}/7)</math>.
 +
 +
Let <math>O1</math> be the center of circle <math>\omega_1</math>  and <math>O2</math> be the center of circle <math>\omega_2</math>. Since circle <math>\omega_1</math> contains both points <math>A</math> and <math>B</math>, <math>O1</math> must lie on the perpendicular bisector of line <math>AB</math>, and similarly <math>O2</math> must lie on the perpendicular bisector of line <math>AC</math>. Through some calculations, we find that the perpendicular bisector of <math>AB</math> has equation <math>x = 3.5</math>, and the perpendicular bisector of <math>AC</math> has equation <math>y = {-11\sqrt{5}/40 \cdot x} + 189\sqrt{5}/80</math>.
 +
 
 +
Since circle <math>\omega_1</math> is tangent to line <math>AC</math> at <math>A</math>, its radius must be perpendicular to <math>AC</math> at <math>A</math>.
 +
Therefore, the radius has equation <math>y = {{-11\cdot\sqrt{5}/40} \cdot x}</math>. Substituting the <math>x</math>-coordinate of <math>O1</math> into this, we find the y-coordinate of <math>O1 = {{-11 \cdot \sqrt{5}/40} \cdot 7/2} = {-77 \cdot \sqrt{5}/80}</math>.
 +
 
 +
Similarly, since circle <math>\omega_2</math> is tangent to line <math>AB</math> at <math>A</math>, its radius must be perpendicular to <math>AB</math> at <math>A</math>. Therefore, the radius has equation <math>x = 0</math> and combining with the previous result for <math>O2</math> we get that the coordinates of <math>O2</math> are <math>(0, 189\sqrt{5}/80)</math>.
 +
 
 +
We now find the slope of <math>O1O2</math>, the line joining the centers of circles <math>\omega_1</math> and <math>\omega_2</math>, which turns out to be <math>{({266 \cdot \sqrt{5} / 80}) \cdot -2/7} = {-19 \cdot \sqrt{5}/20}</math>. Since the <math>y</math>-intercept of that line is at <math>O2(0,189\sqrt{5}/80)</math>, the equation is <math>y = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}</math>.  Since circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>A</math> and <math>K</math>, line <math>AK</math> is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles, <math>AK</math> has slope <math>{4 \cdot \sqrt{5}/19}</math>. Since point <math>A</math> is <math>(0,0)</math>, this line has a <math>y</math>-intercept of <math>0</math>, so it has equation <math>y</math> = <math>{{4 \cdot \sqrt{5}/19} \cdot x}</math>.
 +
 
 +
We set <math>{{4 \cdot \sqrt{5}/19} \cdot x} = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}</math> in order to find the intersection <math>I</math> of the radical axis <math>AK</math> and <math>O1O2</math>. Through some moderate bashing, we find that the intersection point is <math>I(57/28, {3 \cdot \sqrt{5}/7})</math>. 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 <math>A</math> over <math>I</math> yields <math>K</math> and <math>AK</math> = <math>2AI</math> = (This is the most tedious part of the bash) <math>{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</math>. Therefore the answer is <math>9 + 2 = \boxed{011}.</math>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2019|n=II|num-b=10|num-a=12}}
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:52, 28 January 2024

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),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

Supplement

  • 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:

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

~mathboy282

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)

AIME-II-2019-11.png

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 (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=213625"