Difference between revisions of "2015 AIME II Problems/Problem 11"
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==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | unitsize(30); | ||
+ | draw(Circle((0,0),3)); | ||
+ | pair A,B,C,O, Q, P, M, N; | ||
+ | A=(2.5, -sqrt(11/4)); | ||
+ | B=(-2.5, -sqrt(11/4)); | ||
+ | C=(-1.96, 2.28); | ||
+ | Q=(-1.89, 2.81); | ||
+ | P=(1.13, -1.68); | ||
+ | O=origin; | ||
+ | M=foot(O,C,B); | ||
+ | N=foot(O,A,B); | ||
+ | draw(A--B--C--cycle); | ||
+ | label("$A$",A,SE); | ||
+ | label("$B$",B,SW); | ||
+ | label("$C$",C,NW); | ||
+ | label("$Q$",Q,NW); | ||
+ | dot(O); | ||
+ | label("$O$",O,NE); | ||
+ | label("$M$",M,W); | ||
+ | label("$N$",N,S); | ||
+ | label("$P$",P,S); | ||
+ | draw(B--O); | ||
+ | draw(C--Q); | ||
+ | draw(Q--O); | ||
+ | draw(O--C); | ||
+ | draw(O--A); | ||
+ | draw(O--P); | ||
+ | draw(O--M, dashed); | ||
+ | draw(O--N, dashed); | ||
+ | draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5)); | ||
+ | draw(rightanglemark(O,N,B,5)); | ||
+ | draw(rightanglemark(B,O,P,5)); | ||
+ | draw(rightanglemark(O,M,C,5)); | ||
+ | </asy> | ||
+ | |||
+ | Call the <math>M</math> and <math>N</math> foot of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> and let <math>OQ = k</math>. Notice that <math>\triangle{OMB} \sim \triangle{OQB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. Then, <math>MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}</math>. However, since <math>O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math>\frac{r^2}{4.5} = 2 \implies r = 3</math>. We can use the Pythagorean theorem to find <math>ON</math>, so we have <cmath>NB^2 + ON^2 = OB^2 \implies \left(\frac{5}{2}\right)^2 + ON^2 = 3^2 \implies ON = \sqrt{\frac{11}{4}}.</cmath> Likewise, <math>\triangle{PBO} \sim \triangle{PNO}</math> because both are right triangles, and <math>\angle{BPO} \cong \angle{NPO}</math>. Hence, since <math>\triangle{BNO} \sim \triangle{BPO}</math> as well, we have that <math>\triangle{BNO} \sim \triangle{PNO}</math>. It follows that <math>NP = \sqrt{\frac{11}{4}}\left(\frac{\sqrt{{\frac{11}{4}}}}{\frac{5}{2}}\right) = \frac{11}{10}</math>. We add this to <math>BN</math> to get <math>BP</math>, so <math>BP = \frac{5}{2} + \frac{11}{10} = \frac{36}{10} = \frac{18}{5}</math>. Our answer is <math>18 + 5 = \boxed{023}</math>. | ||
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=10|num-a=12}} | {{AIME box|year=2015|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:39, 27 March 2015
Problem
The circumcircle of acute has center . The line passing through point perpendicular to intersects lines and and and , respectively. Also , , , and , where and are relatively prime positive integers. Find .
Solution
Call the and foot of the altitudes from to and , respectively. Let and let . Notice that because both are right triangles, and . Then, . However, since is the circumcenter of triangle , is a perpendicular bisector by the definition of a circumcenter. Hence, . We can use the Pythagorean theorem to find , so we have Likewise, because both are right triangles, and . Hence, since as well, we have that . It follows that . We add this to to get , so . Our answer is .
See also
2015 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.