2017 AMC 10B Problems/Problem 21
Problem
In , , , , and is the midpoint of . What is the sum of the radii of the circles inscribed in and ?
Solution 1
We note that by the converse of the Pythagorean Theorem, is a right triangle with a right angle at . Therefore, , and . Since we have , so the inradius of is , and the inradius of is . Adding the two together, we have .
Solution 2
We have Let be the radius of circle , and let be the radius of circle . We want to find .
We form 6 kites: , , , , , and . Since and are the midpoints of and , respectively, this means that , and .
Since is a kite, , and . The same applies to all kites in the diagram.
Now, we see that , and , thus is , making and isosceles. So, using the Pythagorean Theorem, and also using the Theorem. Hence, we know that .
Notice that the area of the kite (if the opposite angles are right) is , where and denoting each of the 2 congruent sides. This just simplifies to . Hence, we have
and
Solving for and , we find that and , so .
~MrThinker
Solution 3 (Stewart's)
Applying Stewart’s theorem gives us the length of Using that length, we can find the areas of triangles and by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula Therefore, we get Although this solution works perfectly fine, it takes time and has room for error so apply Stewart’s and Heron’s with caution.
~peelybonehead
Video Solution
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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