AMC 10 · 2004 · #22
Grade 8 geometry-2dPick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The sides 5, 12, 13 are a famous Pythagorean triple, so Tool #5 (Look for a Pattern) flags that this is a right triangle before any calculation. That single fact makes both centers easy to pin down, so Tool #1 (Draw a Diagram) sets up a coordinate grid with the right angle at the origin and the legs along the axes. From there Tool #7 (Identify Subproblems) splits the work into three clean pieces: locate the circumcenter, locate the incenter, then measure the straight-line distance between them with the distance formula. Placing everything on coordinates turns 'distance between two special points' into plain arithmetic.
Spot the right triangle
Since 5²+12²=169=13², the converse of the Pythagorean theorem makes this a right triangle, with legs 5 and 12 at the right angle.
When the two shorter sides squared add up to the longest side squared, the corner between them is a perfect right angle.
8.G.B.6Look For A PatternPlace coordinates, find the circumcenter
With C=(0,0), A=(12,0), B=(0,5), the hypotenuse AB is a diameter, so the circumcenter is its midpoint O=(6,5/2).
A right angle always opens onto a diameter, so the hypotenuse's midpoint is exactly the same distance from all three corners.
A right angle always opens onto a diameter, so the hypotenuse's midpoint is the same distance from all three corners.
▸ Why?
All three corners are one radius from that midpoint, which is exactly what a centre means.
▸ Why?
That equal distance is half the hypotenuse, which the right angle guarantees.
Find the inradius and the incenter
Area=1/2·5·12=30 and semiperimeter s=15 give r=30/15=2, and the incircle touches both axes, so the incenter is I=(2,2).
A circle squeezed against both axes must sit one radius away from each, so its center lands at (r,r).
6.G.A.1Identify SubproblemsMeasure the distance between the centers
The gaps are 6-2=4 and 5/2-2=1/2, so the distance formula gives OI=√(16+1/4)=√(65/4)=√(65)/2, choice (D).
Once both centers are points on a grid, the gap between them is just the hypotenuse of the little right triangle formed by their horizontal and vertical offsets.
8.G.B.8Identify SubproblemsIn a right triangle the circumcircle's center sits at the middle of the hypotenuse and the incircle's center sits one inradius in from each leg; drop both points onto a grid and the distance formula finishes the job.
- Spot the right triangle
- Place coordinates, find the circumcenter
- Find the inradius and the incenter
- Measure the distance between the centers