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
Video Solution
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.