Difference between revisions of "2016 AIME I Problems/Problem 15"
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Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>. | Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <math>B</math>, respectively, with line <math>AB</math> closer to point <math>X</math> than to <math>Y</math>. Circle <math>\omega</math> passes through <math>A</math> and <math>B</math> intersecting <math>\omega_1</math> again at <math>D \neq A</math> and intersecting <math>\omega_2</math> again at <math>C \neq B</math>. The three points <math>C</math>, <math>Y</math>, <math>D</math> are collinear, <math>XC = 67</math>, <math>XY = 47</math>, and <math>XD = 37</math>. Find <math>AB^2</math>. | ||
+ | ==Solution== | ||
+ | Using the radical axis theorem, the lines <math>\overline{AD}, \overline{BC}, \overline{XY}</math> are all concurrent at one point, call it <math>F</math>. Now recall by Miquel's theorem in <math>\triangle FDC</math> the fact that quadrilaterals <math>DAXY</math> and <math>CBXY</math> are cyclic implies <math>FAXB</math> is cyclic as well. Denote <math>\omega_{3}\equiv(FAXB)</math> and <math>Z\equiv\ell\cap\overline{FXY}</math>. | ||
+ | |||
+ | Since point <math>Z</math> lies on the radical axis of <math>\omega_{1},\omega_{2}</math>, it has equal power with respect to both circles, thus <cmath>AZ^{2}=\text{Pow}_{\omega_{1}}(Z)=ZX\cdot ZY=\text{Pow}_{\omega_{2}}(Z)=ZB^{2}\implies AZ=ZB.</cmath> Also, notice that <cmath>AZ\cdot ZB=\text{Pow}_{\omega_{3}}(Z)=ZX\cdot ZF\implies ZY=ZF.</cmath> The diagonals of quadrilateral <math>FAYB</math> bisect each other at <math>Z</math>, so we conclude that <math>FAYB</math> is a parallelogram. Let <math>u:=ZX</math>, so that <math>ZY=ZF=u+47</math>. | ||
+ | |||
+ | Because <math>FAYB</math> is a parallelogram and quadrilaterals <math>DAXY, CBXY</math> are cyclic, <cmath>\angle DFX=\angle AFX=\angle BYX=\angle BCX=\angle FCX~~\text{and}~~\angle XDF=\angle XDA=\angle XYA=\angle XFB=\angle XFC</cmath> so we have the pair of similar triangles <math>\triangle DFX~\sim~\triangle FCX</math>. Thus <cmath>\dfrac{37}{2u+47}=\dfrac{2u+47}{67}\implies 2u+47=\sqrt{37\cdot 67}\implies u=\dfrac{1}{2}\left(\sqrt{37\cdot 67}-47\right).</cmath> Now compute <cmath>AB^{2}=4AZ^{2}=4\cdot ZX\cdot ZY=4u(u+47)=37\cdot 67-47^{2}=\textbf{270}.</cmath> | ||
+ | |||
+ | [[File:AIME 2016-I15 Geogebra Diagram.png|840px]] | ||
==Solution 1== | ==Solution 1== | ||
− | Let <math>Z = XY \cap AB</math>. By the radical axis theorem <math>AD, XY, BC</math> are concurrent, say at <math>P</math>. Moreover, <math>\triangle DXP \sim \triangle PXC</math> by simple angle chasing. Let <math>y = PX, x = | + | Let <math>Z = XY \cap AB</math>. By the radical axis theorem <math>AD, XY, BC</math> are concurrent, say at <math>P</math>. Moreover, <math>\triangle DXP \sim \triangle PXC</math> by simple angle chasing. Let <math>y = PX, x = XZ</math>. Then <cmath>\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.</cmath> Now, <math>AZ^2 = \tfrac 14 AB^2</math>, and by power of a point, <cmath>\begin{align*} |
x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ | x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ | ||
x(47+x) &= \tfrac 14 AB^2 | x(47+x) &= \tfrac 14 AB^2 | ||
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==Solution 6 (No words)== | ==Solution 6 (No words)== | ||
− | [[File:2016 AIME I 15.png| | + | [[File:2016 AIME I 15.png|600px|right]] |
− | [[File:2016 AIME I 15b.png| | + | [[File:2016 AIME I 15b.png|600px|left]] |
− | |||
− | ''' | + | <math>AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.</math> |
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 7 (Linearity of Power of a Point)== | ||
+ | Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>P</math>. Let <math>M</math> be the midpoint of segment <math>AB</math>. Then by radical axis on <math>(ADY)</math>, <math>(BCY)</math> and <math>(ABCD)</math>, <math>P</math> lies on <math>XY</math>. By the bisector lemma, <math>M</math> lies on <math>XY</math>. It is well-known that <math>P</math>, <math>A</math>, <math>X</math>, and <math>B</math> are concyclic. By Power of a point on <math>M</math> with respect to <math>(PAXB)</math> and <math>(ADY)</math>, <cmath> |\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY, </cmath> so <math>MP=MY</math>. Thus <math>AB</math> and <math>PY</math> bisect each other, so <math>PAYB</math> is a parallelogram. This implies that <cmath> \angle DAY = \angle YBC, </cmath> so by the inscribed angle theorem <math>\overline{XY}</math> bisects <math>\angle DXC</math>. | ||
+ | |||
+ | Claim: <math>AB^2 = DY \cdot YC</math>. | ||
+ | |||
+ | Proof. Define the linear function <math>f(\bullet) := \text{Pow}(\bullet, (ADY)) - \text{Pow}(\bullet, (ABCD))</math>. Since <math>\overline{BY}</math> is parallel to the radical axis <math>\overline{AD}</math> of <math>(ADY)</math> and <math>(ABCD)</math> by our previous parallelism, <math>f(B)=f(Y)</math>. Note that <math>f(B)=AB^2</math> while <math>f(Y)=DY \cdot YC</math>, so we conclude. <math>\square</math> | ||
+ | |||
+ | By Stewart's theorem on <math>\triangle DXC</math>, <math>DY \cdot YC=37 \cdot 67 - 47^2 = 270</math>, so <math>AB^2=\boxed{270}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/qFfgB15fYS8 | ||
+ | |||
+ | ~r00tsOfUnity | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/QoVIorvv_I8 | https://youtu.be/QoVIorvv_I8 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==Video Solution by The Power of Logic== | ||
+ | https://youtu.be/lTZx6tp2Fvg | ||
==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}} |
Latest revision as of 16:12, 8 January 2024
Contents
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
Using the radical axis theorem, the lines are all concurrent at one point, call it . Now recall by Miquel's theorem in the fact that quadrilaterals and are cyclic implies is cyclic as well. Denote and .
Since point lies on the radical axis of , it has equal power with respect to both circles, thus Also, notice that The diagonals of quadrilateral bisect each other at , so we conclude that is a parallelogram. Let , so that .
Because is a parallelogram and quadrilaterals are cyclic, so we have the pair of similar triangles . Thus Now compute
Solution 1
Let . By the radical axis theorem are concurrent, say at . Moreover, by simple angle chasing. Let . Then Now, , and by power of a point, Solving, we get
Solution 2
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 3
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,
Solution 4
First of all, since quadrilaterals and are cyclic, we can let , and , due to the properties of cyclic quadrilaterals. In addition, let and . Thus, and . Then, since quadrilateral is cyclic as well, we have the following sums: Cancelling out in the second equation and isolating yields . Substituting back into the first equation, we obtain Since we can then imply that . Similarly, . So then , so since we know that bisects , we can solve for and with Stewart’s Theorem. Let and . Then Now, since and , . From there, let and . From angle chasing we can derive that and . From there, since , it is quite clear that , and can be found similarly. From there, since and , we have similarity between , , and . Therefore the length of is the geometric mean of the lengths of and (from ). However, yields the proportion ; hence, the length of is the geometric mean of the lengths of and . We can now simply use arithmetic to calculate .
-Solution by TheBoomBox77
Solution 5 (not too different)
Let . By Radical Axes, lies on . Note that is cyclic as is the Miquel point of in this configuration.
Claim. Proof. We angle chase. and
Let . Note andBy our claim, andFinally, ~Mathscienceclass
Solution 6 (No words)
vladimir.shelomovskii@gmail.com, vvsss
Solution 7 (Linearity of Power of a Point)
Extend and to meet at point . Let be the midpoint of segment . Then by radical axis on , and , lies on . By the bisector lemma, lies on . It is well-known that , , , and are concyclic. By Power of a point on with respect to and , so . Thus and bisect each other, so is a parallelogram. This implies that so by the inscribed angle theorem bisects .
Claim: .
Proof. Define the linear function . Since is parallel to the radical axis of and by our previous parallelism, . Note that while , so we conclude.
By Stewart's theorem on , , so .
~ Leo.Euler
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution
~MathProblemSolvingSkills.com
Video Solution by The Power of Logic
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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