Difference between revisions of "2015 AIME II Problems/Problem 11"
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draw(rightanglemark(O,M,C,5)); | draw(rightanglemark(O,M,C,5)); | ||
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===Solution 1=== | ===Solution 1=== | ||
− | Call | + | Call <math>M</math> and <math>N</math> the feet of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{QOB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. By <math>\frac{MB}{BO}=\frac{BO}{BQ}</math>, <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 BNO</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 (fastest)=== |
− | + | Minor arc <math>BC = 2A</math> so <math>\angle{BOC}=2A</math>. Since <math>\triangle{BOC}</math> is isosceles (<math>BO</math> and <math>OC</math> are radii), <math>\angle{CBO}=(180-2A)/2=90-A</math>. <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>. | |
===Solution 3=== | ===Solution 3=== | ||
− | Let <math>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, < | + | Let <math>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <cmath>OM=\sqrt{r^2-4}</cmath> Thus, <cmath>OQ=\frac{\sqrt{4r^2+9}}{2}</cmath> Using <math>\triangle{BOQ}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. After some calculations with <math>\triangle{BON}</math> ~ <math>\triangle{OPN}</math>, <math>{NP=11/10}</math>. Therefore, <cmath>BP=\frac{5}{2}+\frac{11}{10}=18/5</cmath> <math>18+5=\boxed{023}</math>. |
===Solution 4=== | ===Solution 4=== | ||
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===Solution 5=== | ===Solution 5=== | ||
− | Note: This is not a very good solution, but it is relatively natural and requires next to no thinking. | + | <math>\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}</math> |
Denote the circumradius of <math>ABC</math> to be <math>R</math>, the circumcircle of <math>ABC</math> to be <math>O</math>, and the shortest distance from <math>Q</math> to circle <math>O</math> to be <math>x</math>. | Denote the circumradius of <math>ABC</math> to be <math>R</math>, the circumcircle of <math>ABC</math> to be <math>O</math>, and the shortest distance from <math>Q</math> to circle <math>O</math> to be <math>x</math>. |
Latest revision as of 01:15, 14 January 2022
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 .
Diagram
Solution
Solution 1
Call and the feet of the altitudes from to and , respectively. Let . Notice that because both are right triangles, and . By , . 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 (fastest)
Minor arc so . Since is isosceles ( and are radii), . , so . From this we get that . So , plugging in the given values we get , so , and .
Solution 3
Let . Drawing perpendiculars, and . From there, Thus, Using , we get . Now let's find . After some calculations with ~ , . Therefore, .
Solution 4
Let . Extend to touch the circumcircle at a point . Then, note that . But since is a diameter, , implying . It follows that is a cyclic quadrilateral.
Let . By Power of a Point, The answer is .
Solution 5
Denote the circumradius of to be , the circumcircle of to be , and the shortest distance from to circle to be .
Using Power of a Point on relative to circle , we get that . Using Pythagorean Theorem on triangle to get . Subtracting the first equation from the second, we get that and therefore . Now, set . Using law of cosines on to find in terms of and plugging that into the extended law of sines, we get . Squaring both sides and cross multiplying, we get . Now, we get using quadratic formula. If you drew a decent diagram, is acute and therefore (You can also try plugging in both in the end and seeing which gives a rational solution). Note that Using the cosine addition formula and then plugging in what we know about , we get that . Now, the hard part is to find what is. We therefore want . For the numerator, by inspection will not work for integers and . The other case is if there is . By inspection, works. Therefore, plugging all this in yields the answer, . Solution by hyxue
Solution 6
Reflect , across to points and , respectively with on the circle and collinear. Now, by parallel lines. From here, as collinear. From here, is cyclic, and by power of a point we obtain . ~awang11's sol
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.