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>. | ||
− | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
− | Let <math>Z = XY \cap AB</math>. By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. Furthermore, by simple angle chasing, <math>\triangle DXE \sim \triangle EXC</math>. Let <math>y = EX, x = XZ</math>. Then <math>\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67</math>. Now, by Power of a Point, <math>AZ^2 = | + | Let <math>Z = XY \cap AB</math>. By the Radical Axis Theorem <math>AD, XY, BC</math> concur at point <math>E</math>. Furthermore, by simple angle chasing, <math>\triangle DXE \sim \triangle EXC</math>. Let <math>y = EX, x = XZ</math>. Then <math>\frac{y}{37} = \frac{67}{y} \implies y^2 = 37 \cdot 67</math>. Now, by Power of a Point, <math>AZ^2 = \frac{AB^2}{4}</math>, <math>(y-x)x = \frac{AB^2}{4}</math>, and <math>x(47+x) = \frac{AB^2}{4}</math>. Solving, we get <math>\dfrac{AB ^ 2}{4} = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) \implies AB ^ 2 = 37\cdot67 - 47^2 = \boxed{270}</math> |
===Solution 2=== | ===Solution 2=== | ||
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<cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | <cmath>AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.</cmath> | ||
− | ==Solution | + | ===Solution 4=== |
<asy> | <asy> | ||
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'''-Solution by TheBoomBox77''' | '''-Solution by TheBoomBox77''' | ||
+ | ===Solution 5 (not too different)=== | ||
+ | Let <math>E = DA \cap CB</math>. By Radical Axes, <math>E</math> lies on <math>XY</math>. Note that <math>EAXB</math> is cyclic as <math>X</math> is the Miquel point of <math>\triangle EDC</math> in this configuration. | ||
+ | |||
+ | Claim. <math>\triangle DXE \sim \triangle EXC</math> | ||
+ | Proof. We angle chase. <cmath>\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE</cmath>and<cmath>\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square</cmath> | ||
+ | |||
+ | Let <math>F = EX \cap AB</math>. Note <cmath>FA^2 = FX \cdot FY = FB^2</cmath>and<cmath>EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY</cmath>By our claim, <cmath>\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}</cmath>and<cmath>FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}</cmath>Finally, <cmath>AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare</cmath>~Mathscienceclass | ||
==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 21:07, 10 August 2020
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 .
Solutions
Solution 1
Let . By the Radical Axis Theorem concur at point . Furthermore, by simple angle chasing, . Let . Then . Now, by Power of a Point, , , and . 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
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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