AMC 10 · 2004 · #22
Grade 8 geometry-2dA 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?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A triangle has sides $5$, $12$, and $13$. It has one circle drawn inside it touching all three sides (the inscribed circle) and one circle drawn around it passing through all three vertices (the circumscribed circle). Find the distance between the two circles' centers.
Givens: The triangle's side lengths are $5$, $12$, and $13$.; The inscribed circle (incircle) touches all three sides from inside; its center is the incenter.; The circumscribed circle (circumcircle) passes through all three vertices; its center is the circumcenter.; Answer choices: (A) $\frac{3\sqrt{5}}{2}$, (B) $\frac{7}{2}$, (C) $\sqrt{15}$, (D) $\frac{\sqrt{65}}{2}$, (E) $\frac{9}{2}$.
Unknowns: The distance between the incenter and the circumcenter.
Understand
Restated: A triangle has sides $5$, $12$, and $13$. It has one circle drawn inside it touching all three sides (the inscribed circle) and one circle drawn around it passing through all three vertices (the circumscribed circle). Find the distance between the two circles' centers.
Givens: The triangle's side lengths are $5$, $12$, and $13$.; The inscribed circle (incircle) touches all three sides from inside; its center is the incenter.; The circumscribed circle (circumcircle) passes through all three vertices; its center is the circumcenter.; Answer choices: (A) $\frac{3\sqrt{5}}{2}$, (B) $\frac{7}{2}$, (C) $\sqrt{15}$, (D) $\frac{\sqrt{65}}{2}$, (E) $\frac{9}{2}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
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.
Execute — Answer: D
8.G.B.6 Step 1 Spot the right triangle
- Check the side lengths against the Pythagorean theorem.
- Squaring the two shorter sides gives $5^2+12^2=25+144=169$, and the longest side satisfies $13^2=169$.
- Since $5^2+12^2=13^2$, the converse of the Pythagorean theorem says the triangle is right-angled, with the right angle between the legs of length $5$ and $12$ and the hypotenuse of length $13$ opposite it.
💡 When the two shorter sides squared add up to the longest side squared, the corner between them is a perfect right angle.
6.NS.C.8 Step 2 Place coordinates, find the circumcenter
- Put the right angle at the origin and run the legs along the axes: $C=(0,0)$, $A=(12,0)$, and $B=(0,5)$.
- Then $CA=12$, $CB=5$, and the hypotenuse $AB=\sqrt{12^2+5^2}=13$, matching the triangle.
- In a right triangle the hypotenuse is a diameter of the circumcircle, so the circumcenter $O$ sits at the midpoint of the hypotenuse.
- Averaging the endpoints of $AB$ gives $O=\left(\frac{12+0}{2},\frac{0+5}{2}\right)=\left(6,\tfrac{5}{2}\right)$.
💡 A right angle always opens onto a diameter, so the hypotenuse's midpoint is exactly the same distance from all three corners.
6.G.A.1 Step 3 Find the inradius and the incenter
- The incircle touches all three sides, so its radius $r$ comes from the identity $\text{Area}=r\cdot s$, where $s$ is the semiperimeter.
- The area of the right triangle is $\frac{1}{2}\cdot 5\cdot 12=30$, and the semiperimeter is $s=\frac{5+12+13}{2}=15$, so $r=\frac{\text{Area}}{s}=\frac{30}{15}=2$.
- (The right-triangle shortcut $r=\frac{5+12-13}{2}=2$ agrees.) Because the incircle also touches the two legs lying on the axes, its center is one radius in from each axis, at $I=(r,r)=(2,2)$.
💡 A circle squeezed against both axes must sit one radius away from each, so its center lands at $(r,r)$.
8.G.B.8 Step 4 Measure the distance between the centers
- Now just find the straight-line distance from $O=\left(6,\tfrac{5}{2}\right)$ to $I=(2,2)$ using the distance formula (the Pythagorean theorem in coordinates).
- The horizontal gap is $6-2=4$ and the vertical gap is $\tfrac{5}{2}-2=\tfrac{1}{2}$, so the distance is $\sqrt{4^2+\left(\tfrac12\right)^2}=\sqrt{16+\tfrac14}=\sqrt{\tfrac{65}{4}}=\frac{\sqrt{65}}{2}$.
- The answer is (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.6 Check the side lengths against the Pythagorean theorem. Squaring the two shorter 6.NS.C.8 Put the right angle at the origin and run the legs along the axes: $C=(0,0)$, $A 6.G.A.1 The incircle touches all three sides, so its radius $r$ comes from the identity 8.G.B.8 Now just find the straight-line distance from $O=\left(6,\tfrac{5}{2}\right)$ to Review
Reasonableness: As a decimal $\frac{\sqrt{65}}{2}\approx\frac{8.06}{2}\approx 4.03$. The incenter lies inside the triangle and the circumradius is $R=\frac{13}{2}=6.5$, so the two centers should be less than $6.5$ apart — and $4.03<6.5$ fits. Comparing the choices numerically, (A) $\approx3.35$, (B) $=3.5$, (C) $\approx3.87$, (D) $\approx4.03$, (E) $=4.5$; only (D) matches the computed value, so the answer is stable.
Alternative: Skip coordinates entirely and use Euler's triangle theorem, which relates the distance $d$ between incenter and circumcenter to the two radii: $d^2=R^2-2Rr=R(R-2r)$. Here $R=\frac{13}{2}$ (half the hypotenuse) and $r=2$, so $d^2=\frac{13}{2}\left(\frac{13}{2}-4\right)=\frac{13}{2}\cdot\frac{5}{2}=\frac{65}{4}$, giving $d=\frac{\sqrt{65}}{2}$ — the same answer with no grid.
CCSS standards used (min grade 8)
8.G.B.6Explain a proof of the Pythagorean theorem and its converse (Using $5^2+12^2=13^2$ (the converse) to recognize the $5$-$12$-$13$ triangle as right-angled.)6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Placing the triangle on a coordinate grid and locating the circumcenter at the hypotenuse's midpoint $\left(6,\tfrac{5}{2}\right)$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area $30$ of the right triangle to get the inradius $r=\frac{\text{Area}}{s}=2$ and the incenter $(2,2)$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Measuring the distance $\frac{\sqrt{65}}{2}$ between the incenter $(2,2)$ and the circumcenter $\left(6,\tfrac{5}{2}\right)$.)
⭐ In 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.
⭐ In 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.
More like this
Same archetype — closest grade level first.