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>. 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>. | 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>. | ||
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+ | ==Solution 2== | ||
+ | Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\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 20:47, 27 March 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, . 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 .
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 | |
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All AIME Problems and Solutions |
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