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2017 AMC 10B Problem 21

2017 AMC 10B Problem 21AMC 10 Step by Step

Name: Date: September 13, 2026

Triangle Centers & Cevians · difficulty 3 of 5 · about 5 min · position P21-25

In $\triangle ABC$ , $AB=6$ , $AC=8$ , $BC=10$ , and $D$ is the midpoint of $\overline{BC}$ . What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$ ?

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

Answer

Key insight

The median to the hypotenuse equals half of it, so AD = 5; each half-triangle has area 12, and r = Area/semiperimeter gives 12/8 + 12/9.

Solution

Since $6^2 + 8^2 = 10^2$ , the triangle is right-angled at $A$ with area $\frac12 \cdot 6 \cdot 8 = 24$ .

The midpoint $D$ of the hypotenuse is the circumcenter, so $AD = BD = CD = 5$ . The median $AD$ splits the triangle into two triangles of equal area, $12$ each.

Use $r = \dfrac{\text{Area}}{\text{semiperimeter}}$ for each piece.

$\triangle ADB$ has sides $6, 5, 5$ : semiperimeter $8$ , so $r_1 = \frac{12}{8} = \frac32$ .
$\triangle ADC$ has sides $8, 5, 5$ : semiperimeter $9$ , so $r_2 = \frac{12}{9} = \frac43$ .

Sum: $\frac32 + \frac43 = \frac{9}{6} + \frac{8}{6} = \frac{17}{6}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{17}{6}}$ .

Why this works

Two facts do all the work: the median to the hypotenuse of a right triangle is half the hypotenuse, and any median bisects the area. With all three side lengths and the area of each sub-triangle known, $r = A/s$ is immediate. Whenever a problem asks for an inradius, reach for $A = rs$ before anything else.

Alternative approach

Compute each inradius from scratch: $\triangle ADB$ is isosceles with base $6$ and legs $5$ , so its height is $4$ and area $12$ ; $\triangle ADC$ has base $8$ , legs $5$ , height $3$ , area $12$ . Then apply $r = A/s$ as above. This confirms the areas without invoking the median-bisects-area fact.

The trap

Using r = Area/perimeter instead of Area/semiperimeter, or assuming the two half-triangles have equal inradii because they have equal areas.

Common mistakes

Using r = Area/perimeter instead of Area/semiperimeter, or assuming the two half-triangles have equal inradii because they have equal areas.
Forgetting that $AD = 5$ and trying to compute the median with Stewart's theorem or coordinates, wasting time.

Techniques

Set up the equation/formula and compute; no special trick needed · Cut the figure into known shapes (triangles, rectangles, sectors)