Difference between revisions of "2019 AIME II Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Triangle <math>ABC</math> has side lengths <math>AB=7, BC=8, </math> and <math>CA=9.</math> Circle <math>\omega_1</math> passes through <math>B</math> and is tangent to line <math>AC</math> at <math>A.</math> Circle <math>\omega_2</math> passes through <math>C</math> and is tangent to line <math>AB</math> at <math>A.</math> Let <math>K</math> be the intersection of circles <math>\omega_1</math> and <math>\omega_2</math> not equal to <math>A.</math> Then <math>AK= | + | Triangle <math>ABC</math> has side lengths <math>AB=7, BC=8, </math> and <math>CA=9.</math> Circle <math>\omega_1</math> passes through <math>B</math> and is tangent to line <math>AC</math> at <math>A.</math> Circle <math>\omega_2</math> passes through <math>C</math> and is tangent to line <math>AB</math> at <math>A.</math> Let <math>K</math> be the intersection of circles <math>\omega_1</math> and <math>\omega_2</math> not equal to <math>A.</math> Then <math>AK=\tfrac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
− | ==Solution== | + | ==Solution 1== |
− | 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>. | + | <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 <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>. | ||
-franchester | -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 <math>BK</math> and <math>CK</math> yet we know <math>BC</math> 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 <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{11AK^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) == | ||
+ | |||
+ | 14. 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 AO_{1}B = \angle AO_{2}C = 2x</cmath> by the angle by by tangent. Then we also know that <cmath>\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x</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{OB_{1}}{\sin{90^{\circ}-x}}</cmath> <cmath>\implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}}</cmath> <cmath>\implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>OA_{1} =\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\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}</cmath> <cmath>\implies h =\frac{9}{4}</cmath> Thus our desired length is <math>\frac{9}{2} \implies n+n = \boxed{11}.</math> | ||
+ | |||
+ | ==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> | ||
==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 13:47, 14 October 2021
Contents
Problem
Triangle has side lengths and Circle passes through and is tangent to line at Circle passes through and is tangent to line at Let be the intersection of circles and not equal to Then where and are relatively prime positive integers. Find
Solution 1
-Diagram by Brendanb4321
Note that from the tangency condition that the supplement of with respects to lines and are equal to and , respectively, so from tangent-chord, Also note that , so . Using similarity ratios, we can easily find However, since and , we can use similarity ratios to get
- Now we use Law of Cosines on : From reverse Law of Cosines, . This gives us so our answer is .
-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 and yet we know 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 and radius . Then, we have , or . Similarly, . Notice that is a parallelogram, since and are tangent to and , respectively. Thus, . Now, we get that so by Law of Cosines on we have Then, our answer is . -brianzjk
Solution 3 (Death By Trig Bash)
14. Let the centers of the circles be and where the has the side length contained in the circle. Now let This implies by the angle by by tangent. Then we also know that Now we first find We use law of cosines on to obtain Then applying law of sines on we obtain Using similar logic we obtain
Now we know that Thus using law of cosines on yields While this does look daunting we can write the above expression as Then factoring yields The area Now is twice the length of the altitude of so we let the altitude be and we have Thus our desired length is
Solution 4 (Video)
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI
Solution 5 (Olympiad Geometry)
By the definition of , it is the spiral center mapping , which means that it is the midpoint of the -symmedian chord. In particular, if is the midpoint of and is the reflection of across , we have . By Stewart's Theorem, it then follows that
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.