AMC 10 · 2017 · #21

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
pythagorean-theoreminteger-pythagorean-triplesarea-trianglescoordinate-geometry identify-subproblemsformula-substitution ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
Triangle ABC has sides AB=6, AC=8, and BC=10, and D is the midpoint of side BC. Drawing segment AD splits the triangle into △ ADB and △ ADC. Each smaller triangle has its own inscribed circle. Find the sum of the two inscribed circles' radii.

Pick an answer.

(A)
$\sqrt{5}$
(B)
$\frac{11}{4}$
(C)
$2\sqrt{2}$
(D)
$\frac{17}{6}$
(E)
3

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram): this is a shape problem, and sketching △ ABC with the cevian AD makes the right angle and the two smaller triangles visible. Tool #7 (Identify Subproblems): the inscribed-circle radius of each small triangle comes from its own area and semiperimeter, so the work splits into find AD, find each area, then find each radius. Tool #4 (Introduce a Variable): placing the figure on coordinates lets me compute AD cleanly with the distance formula instead of quoting a memorized theorem.

1STEP 1

Spot the right angle

Since 6²+8²=100=10², the converse of the Pythagorean theorem makes △ ABC a right triangle with the right angle at A.

6²+8²=100=10² → ∠ A=90°
2STEP 2

Find AD with coordinates

Anchor the right angle at the origin: A=(0,0), B=(6,0), C=(0,8), so midpoint D=(3,4) and AD=√(3²+4²)=5, matching BD and DC.

D=(3,4), AD=√(3²+4²)=5
3STEP 3

Split the area in two

With ∠ A=90°, legs 6 and 8 give [△ ABC]=1/2· 6· 8=24; the median to D splits it into two equal-area pieces, each 12.

[△ ABC]=1/2(6)(8)=24, [△ ADB]=[△ ADC]=12
4STEP 4

Get each inscribed radius

Using r=Area/s: △ ADB has s=(6+5+5)/2=8, so r₁=12/8=3/2; △ ADC has s=(8+5+5)/2=9, so r₂=12/9=4/3.

r₁=12/8=3/2, r₂=12/9=4/3
5STEP 5

Add the radii

Over a common denominator of 6: 3/2+4/3=9/6+8/6=17/6, choice (D).

r₁+r₂=3/2+4/3=9/6+8/6=17/6 → (D)
Answer
17/6
The sum 17/6≈ 2.83 should sit between the two individual radii doubled and a sanity bound. Each radius is under 5/2 (the radii are 1.5 and about 1.33), so their sum near 2.83 is believable for two small triangles whose biggest side is only 8. The answer is also positive and not a clean integer, consistent with the mixed side lengths 6,8. Choices like √5≈2.24 or 3 bracket it, so 17/6 landing between them is reasonable.
💡Key takeaway

An inscribed circle's radius is area over semiperimeter, so once you see 6,8,10 is a right triangle and AD=5, the two radii 3/2 and 4/3 add to 17/6, choice (D).

  • Spot the right angle
  • Find AD with coordinates
  • Split the area in two
  • Get each inscribed radius
  • Add the radii

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