Difference between revisions of "2016 AIME I Problems/Problem 15"
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<cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath> | <cmath>37n \cdot 67n + 47^2 = 37 \cdot 67</cmath> | ||
<cmath>n^2 = \frac{270}{2479}</cmath> | <cmath>n^2 = \frac{270}{2479}</cmath> | ||
− | Now, since <math>\angle AYX = x</math> and <math>\angle BYX = y</math>, <math>\angle AYB = x + y</math>. From there, let <math>\angle AYD = \alpha</math> and <math>\angle BYC = \beta</math>. From angle chasing we can derive that <math>\angle YDX = \angle YAX = \beta - x</math> and <math>\angle YCX = \angle YBX = \alpha - y</math>. From there, since <math>\angle ADX = x</math>, it is quite clear that <math>\angle ADY = \beta</math>, and <math>\angle YAB = \beta</math> can be found similarly. From there, since <math>\angle ADY = \angle YAB = \angle BYC = \beta</math> and <math>\angle DAY = \angle AYB = \angle YBC = x + y</math>, we have <math>AA</math> similarity between <math>\triangle DAY</math>, <math>\triangle AYB</math>, and <math>\triangle YBC</math>. Therefore the length of <math>AY</math> is the geometric mean of the lengths of <math>DA</math> and <math>YB</math> (from <math>\triangle DAY \sim \triangle AYB</math>). However, <math>\triangle DAY \sim \triangle AYB \sim \triangle YBC</math> yields the proportion <math>\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}</math>; | + | Now, since <math>\angle AYX = x</math> and <math>\angle BYX = y</math>, <math>\angle AYB = x + y</math>. From there, let <math>\angle AYD = \alpha</math> and <math>\angle BYC = \beta</math>. From angle chasing we can derive that <math>\angle YDX = \angle YAX = \beta - x</math> and <math>\angle YCX = \angle YBX = \alpha - y</math>. From there, since <math>\angle ADX = x</math>, it is quite clear that <math>\angle ADY = \beta</math>, and <math>\angle YAB = \beta</math> can be found similarly. From there, since <math>\angle ADY = \angle YAB = \angle BYC = \beta</math> and <math>\angle DAY = \angle AYB = \angle YBC = x + y</math>, we have <math>AA</math> similarity between <math>\triangle DAY</math>, <math>\triangle AYB</math>, and <math>\triangle YBC</math>. Therefore the length of <math>AY</math> is the geometric mean of the lengths of <math>DA</math> and <math>YB</math> (from <math>\triangle DAY \sim \triangle AYB</math>). However, <math>\triangle DAY \sim \triangle AYB \sim \triangle YBC</math> yields the proportion <math>\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}</math>; hence, the length of <math>AB</math> is the geometric mean of the lengths of <math>DY</math> and <math>YC</math>. |
We can now simply use arithmetic to calculate <math>AB^2</math>. | We can now simply use arithmetic to calculate <math>AB^2</math>. | ||
<cmath>AB^2 = DY \cdot YC</cmath> | <cmath>AB^2 = DY \cdot YC</cmath> |
Revision as of 23:20, 19 October 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,
Solution 3
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.