Difference between revisions of "2016 AIME I Problems/Problem 15"
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===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 = \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>AB^2 = \left(\frac{y - 47}{2}\right)\left(\frac{y + 47}{2}\right) = 37\cdot67 - 47^2 = \boxed{270}</math> | + | 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) = 37\cdot67 - 47^2 = \boxed{270}</math> |
===Solution 2=== | ===Solution 2=== |
Revision as of 11:13, 2 September 2019
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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.