Difference between revisions of "2019 AIME II Problems/Problem 11"
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Then, our answer is <math>9+2=\boxed{11}</math>. | Then, our answer is <math>9+2=\boxed{11}</math>. | ||
-brianzjk | -brianzjk | ||
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+ | == Solution 3 (Death By Trig Bash) == | ||
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+ | 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 <math>\angle AO_{1}B = \angle AO_{2}C = 2x</math> by the angle by by tangent. Then we also know that <math>\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x.</math> Now we first find <math>\cos x.</math> We use law of cosines on <math>\bigtriangleup ABC</math> to obtain <math>64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x} \implies \cos{x} = \displaystyle\frac{11}{21} \implies \sin{x} = \displaystyle\frac{8\sqrt{5}}{21}.</math> Then applying law of sines on <math>\bigtriangleup AO_{1}B</math> we obtain <math>\displaystyle\frac{7}{\sin{2x}} = \displaystyle\frac{OB_{1}}{\sin{90^{\circ}-x}} \implies \displaystyle\frac{7}{2\sin{x}\cos{x}} = \displaystyle\frac{OB_{1}}{\cos{x}} \implies OB_{1} = O_{1}A= \displaystyle\frac{147}{16\sqrt{5}}.</math> Using similar logic we obtain <math>OA_{1} = \displaystyle\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 <math>O_{1}O_{2} = \displaystyle\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}}.</math> While this does look daunting we can write the above expression as <math>\displaystyle\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}} = \displaystyle\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}.</math> Then factoring yields <math>\sqrt{\displaystyle\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} = \displaystyle\frac{147}{8\sqrt{5}}.</math> The area <math>[O_{1}AO_{2}] = \displaystyle\frac{1}{2} \cdot \displaystyle\frac{147}{16\sqrt{5}} \cdot \displaystyle\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) = \displaystyle\frac{1}{2} \cdot \displaystyle\frac{147}{16\sqrt{5}} \cdot \displaystyle\frac{189}{16\sqrt{5}} \cdot \displaystyle\frac{8\sqrt{5}}{21}.</math> 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 <math>h \cdot \displaystyle\frac{147\sqrt{5}}{8\sqrt{5}} = \displaystyle\frac{1}{2} \cdot \displaystyle\frac{147}{16\sqrt{5}} \cdot \displaystyle\frac{189}{16\sqrt{5}} \cdot \displaystyle\frac{8\sqrt{5}}{21} \implies h = \displaystyle\frac{9}{4}.</math> Thus our desired length is <math>\displaystyle\frac{9}{2} \implies n+n = \boxed{11}.</math> | ||
==Solution 3 (Video)== | ==Solution 3 (Video)== |
Revision as of 19:24, 8 August 2020
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
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 3 (Video)
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI
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 |
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