AMC 10 · 2018 · #21

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

inradiustangent-circlescoordinate-geometryinteger-pythagorean-triplesshoelace-formula coordinate-geometryidentify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremcoordinate-geometryinradius

📏 Long solution 💡 4 insights

Problem

A triangle has sides 13, 12, and 5. Let the circumcentre be the centre of the circle through all three vertices, and the incentre the centre of the circle inside touching all three sides. A third circle touches both legs and also touches the circumcircle from the inside. Find the area of the triangle formed by those three centres.

Pick an answer.

(A)

$\frac52$

(B)

$\frac{11}{4}$

(C)

(D)

$\frac{13}{4}$

(E)

$\frac72$

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram): 5² + 12² = 13², so the angle at C is right — put C at the origin with the two legs on the axes and every center becomes a pair of numbers instead of a description. Tool #7 (Identify Subproblems): the question asks for one area but hands me three unrelated centers, so I locate O, then I, then M as three separate small jobs. Tool #4 (Introduce a Variable): the only thing unknown about ⊙ M is its size, so I name its radius a; tangency to both axes forces its center to be (a,a), one letter carrying the whole circle. Tool #13 (Convert to Algebra): "touches the circumcircle from inside" is a picture, but it says exactly that the distance between the centers equals the difference of the radii — one equation in a.

1STEP 1

Spot the right angle, set coordinates

The sides form a right triangle.

5²+12²=13² → ∠ ACB = 90°; C=(0,0), A=(12,0), B=(0,5)

2STEP 2

Locate the circumcenter

The circumcentre is the hypotenuse's midpoint.

O=((12+0)/2,(0+5)/2)=(6,5/2), R=13/2

3STEP 3

Locate the incenter

Area and perimeter give the inradius.

r=[ABC]/s=30/15=2 → I=(2,2)

4STEP 4

Name the third circle with one letter

The third circle needs only one letter.

M=(a,a) with radius a, a ＞ 0

5STEP 5

Turn internal tangency into an equation

Internal tangency becomes a distance equation.

√((a-6)²+(a-5/2)²)=13/2-a

6STEP 6

Solve for the radius

Solving fixes the third centre.

2a²-17a+169/4=a²-13a+169/4 → a²-4a=0 → a=4, M=(4,4)

7STEP 7

Take the area of triangle MOI

The area is seven halves.

[MOI]=1/2|4(5/2-2)+6(2-4)+2(4-5/2)|=1/2|2-12+3|=7/2 → (E)

Answer

7/2

First check a=4 really works: OM=√((4-6)²+(4-5/2)²)=√(4+9/4)=5/2, and R-a=13/2-4=5/2 — they match, so the circles touch internally. The touch points on the legs are (4,0) and (0,4), which lie on the actual segments since 4 ≤ 12 and 4 ≤ 5, so ⊙ M touches the legs themselves, not their extensions. Now recompute the area a second way. Both I and M have equal coordinates, so both lie on the bisector y=x of the right angle at C; that line is the base. MI=√((4-2)²+(4-2)²)=2√2, and the distance from O(6,5/2) to x-y=0 is (|6-5/2|)/√2=7√2/4, giving [MOI]=1/2 · 2√2·7√2/4=7/2 — the same value. The three centers are genuinely not collinear, since O is off the line y=x, so a positive area is expected. Finally, the choices themselves confirm the tangency was internal: taking OM=R+a instead would give a²-30a=0, so a=30 and an area of 49, nowhere near the list.

💡Key takeaway

A circle touching both arms of a right angle has its center on the 45° line, and touching another circle from the inside just means the two centers are the difference of the radii apart.

Spot the right angle, set coordinates
Locate the circumcenter
Locate the incenter
Name the third circle with one letter
Turn internal tangency into an equation
Solve for the radius
Take the area of triangle MOI