Difference between revisions of "2010 AIME I Problems/Problem 15"
m (→See Also) |
m (→Solution 2) |
||
Line 55: | Line 55: | ||
Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC)/2 = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields | Let the incircle of <math>\triangle ABM</math> meet <math>AM</math> at <math>P</math> and the incircle of <math>\triangle BCM</math> meet <math>CM</math> at <math>Q</math>. Then note that <math>I_1 P Q I_2</math> is a rectangle. Also, <math>\angle I_1 M I_2</math> is right because <math>MI_1</math> and <math>MI_2</math> are the angle bisectors of <math>\angle AMB</math> and <math>\angle CMB</math> respectively and <math>\angle AMB + \angle CMB = 180^\circ</math>. By properties of [[tangent (geometry)|tangents]] to [[circle]]s <math>MP = (MA + MB - AB)/2 = x + y - 6</math> and <math>MQ = (MB + MC - BC)/2 = - x + y + 1</math>. Now notice that the altitude of <math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)</math> (3). Equating (3) with (1) and (2) separately yields | ||
− | <cmath> | + | <cmath>\begin{align*} |
2y^2 - 30 = 2xy + 5x - 7y \\ | 2y^2 - 30 = 2xy + 5x - 7y \\ | ||
− | 2y^2 - 70 = - 2xy - 5x + 7y, | + | 2y^2 - 70 = - 2xy - 5x + 7y, \end{align*} |
</cmath> | </cmath> | ||
Revision as of 11:49, 30 July 2017
Contents
Problem
In with , , and , let be a point on such that the incircles of and have equal radii. Let and be positive relatively prime integers such that . Find .
Solution
Solution 1
Let , then . Also let Clearly, . We can also express each area by the rs formula. Then . Equating and cross-multiplying yields or Note that for d to be positive, we must have .
By Stewart's Theorem, we have or Brute forcing by plugging in our previous result for , we have Clearing the fraction and gathering like terms, we get
Aside: Since must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that is an integer. The only such in the above-stated range is .
Legitimately solving that quartic, note that and should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get The only solution in the desired range is thus . Then , and our desired ratio , giving us an answer of .
Solution 2
Let and so . Let the incenters of and be and respectively, and their equal inradii be . From , we find that
Let the incircle of meet at and the incircle of meet at . Then note that is a rectangle. Also, is right because and are the angle bisectors of and respectively and . By properties of tangents to circles and . Now notice that the altitude of to is of length , so by similar triangles we find that (3). Equating (3) with (1) and (2) separately yields
and adding these we have
Solution 3
Let the incircle of hit , , at , and let the incircle of hit , , at . Draw the incircle of , and let it be tangent to at . Observe that we have a homothety centered at A sending the incircle of to that of , and one centered at taking the incircle of to that of . These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is .
By standard computations, and . Now, let and . We will now go around and chase lengths. Observe that . Then, . We also have , so and .
Observe now that . Also,. Solving, we get and (as a side note, note that , a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get . To finish, we will compute area ratios. . Also, since their inradii are equal, we get . Equating and cross multiplying yields the quadratic , so . However, observe that , so we take . Our ratio is therefore , giving the answer .
Solution 4
Suppose the incircle of touches at , and the incircle of touches at . Then
We have ,
, ,
Therefore
And since , ,
Now,
So or . But from (1) we know that , or , so , , .
Sidenote
In the problem, and the equal inradius of the two triangles happens to be .
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.