Difference between revisions of "2015 AIME II Problems/Problem 11"
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Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that <math>x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}</math>. Using Pythagorean Theorem on triangle <math>QOB</math> to get <math>(x + r)^2 + r^2 = \frac{81}{4}</math>. Subtracting the first equation from the second, we get that <math>2r^2 = 18</math> and therefore <math>r = 3</math>. Now, set <math>\cos{ABC} = y</math>. Using law of cosines on <math>ABC</math> to find <math>AC</math> in terms of <math>y</math> and plugging that into the extended law of sines, we get <math>\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6</math>. Squaring both sides and cross multiplying, we get <math>36x^2 - 40x + 5 = 0</math>. Now, we get <math>x = \frac{10 \pm \sqrt{55}}{18}</math> using quadratic formula. If you drew a decent diagram, <math>B</math> is acute and therefore <math>x = \frac{10 + \sqrt{55}}{18}</math>(You can also try plugging in both in the end and seeing which gives a rational solution). Note that <math>BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.</math> Using the cosine addition formula and then plugging in what we know about <math>QBO</math>, we get that <math>BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}</math>. Now, the hard part is to find what <math>\sin{B}</math> is. We therefore want <math>\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}</math>. For the numerator, by inspection <math>(a + b\sqrt{55})^2</math> will not work for integers <math>a</math> and <math>b</math>. The other case is if there is <math>(a\sqrt{5} + b\sqrt{11})^2</math>. By inspection, <math>5\sqrt{5} - 2\sqrt{11}</math> works. Therefore, plugging all this in yields the answer, <math>\frac{18}{5} \rightarrow \boxed{23}</math>. Solution by hyxue | Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that <math>x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}</math>. Using Pythagorean Theorem on triangle <math>QOB</math> to get <math>(x + r)^2 + r^2 = \frac{81}{4}</math>. Subtracting the first equation from the second, we get that <math>2r^2 = 18</math> and therefore <math>r = 3</math>. Now, set <math>\cos{ABC} = y</math>. Using law of cosines on <math>ABC</math> to find <math>AC</math> in terms of <math>y</math> and plugging that into the extended law of sines, we get <math>\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6</math>. Squaring both sides and cross multiplying, we get <math>36x^2 - 40x + 5 = 0</math>. Now, we get <math>x = \frac{10 \pm \sqrt{55}}{18}</math> using quadratic formula. If you drew a decent diagram, <math>B</math> is acute and therefore <math>x = \frac{10 + \sqrt{55}}{18}</math>(You can also try plugging in both in the end and seeing which gives a rational solution). Note that <math>BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.</math> Using the cosine addition formula and then plugging in what we know about <math>QBO</math>, we get that <math>BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}</math>. Now, the hard part is to find what <math>\sin{B}</math> is. We therefore want <math>\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}</math>. For the numerator, by inspection <math>(a + b\sqrt{55})^2</math> will not work for integers <math>a</math> and <math>b</math>. The other case is if there is <math>(a\sqrt{5} + b\sqrt{11})^2</math>. By inspection, <math>5\sqrt{5} - 2\sqrt{11}</math> works. Therefore, plugging all this in yields the answer, <math>\frac{18}{5} \rightarrow \boxed{23}</math>. Solution by hyxue | ||
+ | |||
+ | ===Solution 6=== | ||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | import graph; size(15cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */ | ||
+ | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
+ | /* draw figures */ | ||
+ | draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); | ||
+ | draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); | ||
+ | draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */ | ||
+ | draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); | ||
+ | /* dots and labels */ | ||
+ | dot((-1.82,0.87),dotstyle); | ||
+ | label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); | ||
+ | dot((3.178984115621537,0.7692140299269852),dotstyle); | ||
+ | label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); | ||
+ | dot((2.6857306099246263,4.738685150758791),dotstyle); | ||
+ | label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); | ||
+ | dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); | ||
+ | label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); | ||
+ | dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); | ||
+ | label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); | ||
+ | dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); | ||
+ | label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); | ||
+ | dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); | ||
+ | label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); | ||
+ | dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); | ||
+ | label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); | ||
+ | dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); | ||
+ | label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | </asy> | ||
+ | Reflect <math>A</math>, <math>P</math> across <math>OB</math> to points <math>A'</math> and <math>P'</math>, respectively with <math>A'</math> on the circle and <math>P, O, P'</math> collinear. Now, <math>\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB</math> by parallel lines. From here, <math>\angle P'PB = \angle PP'B = \angle A'P'Q</math> as <math>P, P', Q</math> collinear. From here, <math>A'P'QC</math> is cyclic, and by power of a point we obtain <math>\frac{18}{5} \implies \boxed{023}</math>. | ||
+ | ~awang11's sol | ||
===Solution 6=== | ===Solution 6=== |
Revision as of 22:21, 6 March 2020
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 1
Call the and foot 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
Notice that , 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
Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.
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
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.