Difference between revisions of "2016 AIME I Problems/Problem 15"
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==Solution== | ==Solution== | ||
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− | Let <math>AB</math> and <math>EY</math> intersect at <math>S</math>. Note that because <math>AXDY</math> and <math>CYXB</math> are cyclic, by Miquel | + | ===Solution 1=== |
+ | By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. | ||
+ | |||
+ | Let <math>AB</math> and <math>EY</math> intersect at <math>S</math>. Note that because <math>AXDY</math> and <math>CYXB</math> are cyclic, by Miquel's Theorem <math>AXBE</math> is cyclic as well. Thus | ||
<cmath>\angle AEX = \angle ABX = \angle XCB = \angle XYB</cmath>and | <cmath>\angle AEX = \angle ABX = \angle XCB = \angle XYB</cmath>and | ||
− | <cmath>\angle XEB = \angle XAB = \angle XDA = \angle XYA.</cmath>Thus <math>AY | + | <cmath>\angle XEB = \angle XAB = \angle XDA = \angle XYA.</cmath>Thus <math>AY \parallel EB</math> and <math>YB \parallel EA</math>, so <math>AEBY</math> is a parallelogram. Hence <math>AS = SB</math> and <math>SE = SY</math>. But notice that <math>DXE</math> and <math>EXC</math> are similar by <math>AA</math> Similarity, so <math>XE^2 = XD \cdot XC = 37 \cdot 67</math>. But |
<cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 37 \cdot 67 - 47^2 = 270.</math> | <cmath>XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.</cmath>Hence <math>AB^2 = 37 \cdot 67 - 47^2 = 270.</math> | ||
+ | |||
+ | |||
+ | ===Solution 2=== | ||
+ | First, we note that as <math>\triangle XDY</math> and <math>\triangle XYC</math> have bases along the same line, <math>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}</math>. We can also find the ratio of their areas using the circumradius area formula. If <math>R_1</math> is the radius of <math>\omega_1</math> and if <math>R_2</math> is the radius of <math>\omega_2</math>, then | ||
+ | <cmath>\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.</cmath> | ||
+ | Since we showed this to be <math>\frac{DY}{YC}</math>, we see that <math>\frac{R_2}{R_1}=\frac{67}{37}</math>. | ||
+ | |||
+ | We extend <math>AD</math> and <math>BC</math> to meet at point <math>P</math>, and we extend <math>AB</math> and <math>CD</math> to meet at point <math>Q</math> as shown below. | ||
+ | <asy> | ||
+ | size(200); | ||
+ | import olympiad; | ||
+ | real R1=45,R2=67*R1/37; | ||
+ | real m1=sqrt(R1^2-23.5^2); | ||
+ | real m2=sqrt(R2^2-23.5^2); | ||
+ | pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); | ||
+ | draw(circle(o1,R1)); | ||
+ | draw(circle(o2,R2)); | ||
+ | pair q=(-R1/(R2-R1)*o2.x,0); | ||
+ | pair a=tangent(q,o1,R1,2); | ||
+ | pair b=tangent(q,o2,R2,2); | ||
+ | pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; | ||
+ | pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; | ||
+ | pair p=extension(a,d,b,c); | ||
+ | dot(q^^a^^b^^x^^y^^c^^d^^p); | ||
+ | draw(q--b^^q--c); | ||
+ | draw(p--d^^p--c^^x--y); | ||
+ | draw(a--y^^b--y); | ||
+ | draw(d--x--c); | ||
+ | label("$A$",a,NW,fontsize(8)); | ||
+ | label("$B$",b,NE,fontsize(8)); | ||
+ | label("$C$",c,SE,fontsize(8)); | ||
+ | label("$D$",d,SW,fontsize(8)); | ||
+ | label("$X$",x,2*WNW,fontsize(8)); | ||
+ | label("$Y$",y,3*S,fontsize(8)); | ||
+ | label("$P$",p,N,fontsize(8)); | ||
+ | label("$Q$",q,W,fontsize(8)); | ||
+ | </asy> | ||
+ | As <math>ABCD</math> is cyclic, we know that <math>\angle BCD=180-\angle DAB=\angle BAP</math>. But then as <math>AB</math> is tangent to <math>\omega_2</math> at <math>B</math>, we see that <math>\angle BCD=\angle ABY</math>. Therefore, <math>\angle ABY=\angle BAP</math>, and <math>BY\parallel PD</math>. A similar argument shows <math>AY\parallel PC</math>. These parallel lines show <math>\triangle PDC\sim\triangle ADY\sim\triangle BYC</math>. Also, we showed that <math>\frac{R_2}{R_1}=\frac{67}{37}</math>, so the ratio of similarity between <math>\triangle ADY</math> and <math>\triangle BYC</math> is <math>\frac{37}{67}</math>, or rather | ||
+ | <cmath>\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.</cmath> | ||
+ | We can now use the parallel lines to find more similar triangles. As <math>\triangle AQD\sim \triangle BQY</math>, we know that | ||
+ | <cmath>\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.</cmath> | ||
+ | Setting <math>QA=37x</math>, we see that <math>QB=67x</math>, hence <math>AB=30x</math>, and the problem simplifies to finding <math>30^2x^2</math>. Setting <math>QD=37^2y</math>, we also see that <math>QY=37\cdot 67y</math>, hence <math>DY=37\cdot 30y</math>. Also, as <math>\triangle AQY\sim \triangle BQC</math>, we find that | ||
+ | <cmath>\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.</cmath> | ||
+ | As <math>QY=37\cdot 67y</math>, we see that <math>QC=67^2y</math>, hence <math>YC=67\cdot30y</math>. | ||
+ | |||
+ | Applying Power of a Point to point <math>Q</math> with respect to <math>\omega_2</math>, we find | ||
+ | <cmath>67^2x^2=37\cdot 67^3 y^2,</cmath> | ||
+ | or <math>x^2=37\cdot 67 y^2</math>. We wish to find <math>AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2</math>. | ||
+ | |||
+ | Applying Stewart's Theorem to <math>\triangle XDC</math>, we find | ||
+ | <cmath>37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).</cmath> | ||
+ | We can cancel <math>30\cdot 104\cdot y</math> from both sides, finding <math>37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2</math>. Therefore, | ||
+ | <cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2016|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2016|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:02, 6 March 2016
Problem
Circles and intersect at points and . Line is tangent to and at and , respectively, with line closer to point than to . Circle passes through and intersecting again at and intersecting again at . The three points , , are collinear, , , and . Find .
Solution
Solution 1
By the Radical Axis Theorem concur at point .
Let and intersect at . Note that because and are cyclic, by Miquel's Theorem is cyclic as well. Thus and Thus and , so is a parallelogram. Hence and . But notice that and are similar by Similarity, so . But Hence
Solution 2
First, we note that as and have bases along the same line, . We can also find the ratio of their areas using the circumradius area formula. If is the radius of and if is the radius of , then Since we showed this to be , we see that .
We extend and to meet at point , and we extend and to meet at point as shown below. As is cyclic, we know that . But then as is tangent to at , we see that . Therefore, , and . A similar argument shows . These parallel lines show . Also, we showed that , so the ratio of similarity between and is , or rather We can now use the parallel lines to find more similar triangles. As , we know that Setting , we see that , hence , and the problem simplifies to finding . Setting , we also see that , hence . Also, as , we find that As , we see that , hence .
Applying Power of a Point to point with respect to , we find or . We wish to find .
Applying Stewart's Theorem to , we find We can cancel from both sides, finding . Therefore,
See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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