Difference between revisions of "2016 AIME I Problems/Problem 15"
m (→Solution 5 (not too different)) |
m (→Solution 5 (not too different)) |
||
Line 116: | Line 116: | ||
'''-Solution by TheBoomBox77''' | '''-Solution by TheBoomBox77''' | ||
===Solution 5 (not too different)=== | ===Solution 5 (not too different)=== | ||
− | Let <math>E = DA \cap CB</math>. By | + | 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> | Claim. <math>\triangle DXE \sim \triangle EXC</math> |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.