2007 AIME II Problems/Problem 15
Problem
Four circles and with the same radius are drawn in the interior of triangle such that is tangent to sides and , to and , to and , and is externally tangent to and . If the sides of triangle are and the radius of can be represented in the form , where and are relatively prime positive integers. Find
Contents
Solution
Solution 1 (Homothety)
First, apply Heron's formula to find that . The semiperimeter is , so the inradius is .
Now consider the incenter of . Let the radius of one of the small circles be . Let the centers of the three little circles tangent to the sides of be , , and . Let the center of the circle tangent to those three circles be . The homothety maps to ; since , is the circumcenter of and therefore maps the circumcenter of to . Thus, , where is the circumradius of . Substituting , and the answer is .
https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png
Solution 2
Consider a 13-14-15 triangle. [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is , where is the semiperimeter. Scale the triangle with the inradius by a linear scale factor,
The circumradius is where and are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, .
Cut and combine the triangles, as shown. Then solve for :
The solution is .
Solution 3 (elementary)
Let , , , and be the centers of circles , , , , respectively, and let be their radius.
Now, triangles and are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for .
Since , is the circumcenter of triangle and its circumradius is . Let denote the incenter of triangle and the inradius of . Then the inradius of , so now we compute r. Computing the inradius by , we find that the inradius of is . Additionally, using the circumradius formula where is the area of and is the circumradius, we find . Now we can equate the ratio of circumradius to inradius in triangles and .
Solving, we get , so our answer is .
Solution 4
According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is. Now denoting , and centers of circles tangent to are relatively with both perpendicular to . It is easy to know that , so according to half angle formula. Similarly, we can find . So we can see that . Obviously, . After solving, we get , so our answer is . ~bluesoul
Sidenote (Generalization)
If four circles and with the same radius are drawn in the interior of triangle such that is tangent to sides and , to and , to and , and is externally tangent to and . If has side lengths and , then the radius of can be written as or, more simply as, where is the area of the triangle and is the semiperimeter
~pinkpig
Solution 5
Let be the centers of , respectively. Also, let be the incenter of and be the radius of circle . Since , , and , we know that
That means , , and . Thus, . We also know that we are scaling each side of (from to for instance), about (since A,O_A,I are collinear; same apply with and ).
Now, let the homothety map to . To start off, we know the circumradius of is , since . Since , , , we can get an relationship involving and via another way to find the circumradius:
Take notice of the inradius of . We get the inradius to be . Let the tangency point of the incircle and side be . We know . We also know that we can cut off the part of that is outside of to get the inradius of . To part that is outside turns out just to be the radius of circle (as seen in the picture). That means the inradius of is just . We can calculate that incradius in another way, though. We know that the inradius of is , which means the inradius of is just (by our homethety ratio).
Thus, we have . That gives as our final answer.
The homethety turned out to be
~sml1809
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
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