Difference between revisions of "2004 AMC 10B Problems/Problem 22"
(Undo revision 164131 by Pfalcon (talk). This solution is a verbatim restatement of Solution 3 but without the helpful diagram.) (Tag: Undo) |
Math-lover1 (talk | contribs) m (→Solution 2: I gave an alternate method to finding the circumradius.) |
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We can see that this is a right triangle, and hence has area <math>30</math>. We then find the inradius with the formula <math>A=rs</math>, where <math>s</math> denotes semiperimeter. We easily see that <math>s=15</math>, so <math>r=2</math>. | We can see that this is a right triangle, and hence has area <math>30</math>. We then find the inradius with the formula <math>A=rs</math>, where <math>s</math> denotes semiperimeter. We easily see that <math>s=15</math>, so <math>r=2</math>. | ||
− | We now find the circumradius with the formula <math>A=\frac{abc}{4R}</math>. Solving for <math>R</math> gives <math>R=\frac{13}{2}</math>. | + | We now find the circumradius with the formula <math>A=\frac{abc}{4R}</math>. Solving for <math>R</math> gives <math>R=\frac{13}{2}</math>. Additionally, we may notice that the side lengths <math>(5, 12, 13)</math> are in a Pythagorean triple, and therefore the triangle is right for a circumradius of <math>\frac{13}{2}</math>. |
Substituting all of this back into our formula gives: | Substituting all of this back into our formula gives: |
Revision as of 13:46, 6 April 2024
Problem
A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?
Solution 1
This is a right triangle. Pick a coordinate system so that the right angle is at
and the other two vertices are at
and
.
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .
The radius of the inscribed circle can be computed using the well-known identity
, where
is the area of the triangle and
its perimeter. In our case,
and
. Thus,
. As the inscribed circle touches both legs, its center must be at
.
The distance of these two points is then .
Solution 2
We directly apply Euler’s Theorem, which states that if the circumcenter is and the incenter
, and the inradius is
and the circumradius is
, then
We can see that this is a right triangle, and hence has area . We then find the inradius with the formula
, where
denotes semiperimeter. We easily see that
, so
.
We now find the circumradius with the formula . Solving for
gives
. Additionally, we may notice that the side lengths
are in a Pythagorean triple, and therefore the triangle is right for a circumradius of
.
Substituting all of this back into our formula gives:
So,
Solution 3
Construct
such that
,
, and
. Since this is a pythagorean triple,
. By a property of circumcircles and right triangles, the circumcenter,
, lies on the midpoint of
, so
. Turning to the incircle, we find that the inradius is
, using the formula
, where
is the area of the triangle,
is the inradius, and
is the semiperimeter. We then denote the incenter
, along with the points of tangency
,
, and
. Because
by a property of tangency,
, and so
is a square. Then, since
,
. As
,
, and because
by HL,
. Therefore,
. Because
, pythagorean theorem gives
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.