Difference between revisions of "2015 AIME II Problems/Problem 11"
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− | 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>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle NBO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math> | + | 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>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle NBO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math>m + n=\boxed{023}</math>.. |
==Solution 2== | ==Solution 2== |
Revision as of 03:20, 9 July 2015
Contents
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, . Since we know and , we have . Thus, . ..
Solution 2
Notice that , so . From this we get that . So , plugging in the given values we get , so , and .
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.