Printable

2017 AMC 10B Problem 21

2017 AMC 10B Problem 21AMC 10 Step by Step

Name: Date: October 28, 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 right angle at A makes AD = 5. Each half-triangle has area 12, and each radius is its area divided by half its perimeter.

Solution

For a circle inside a triangle, the useful fact is: area $=$ radius $\times$ half the perimeter. (Join the circle's center to the three corners. This cuts the triangle into three pieces, each with a side as base and the radius as height. Adding $\frac12 \cdot \text{side} \cdot r$ over the three sides gives $r$ times half the perimeter.) So for each half-triangle we need its sides and its area.

A right triangle. $6^2 + 8^2 = 36 + 64 = 100 = 10^2$ , so the angle at $A$ is a right angle. The area of $ABC$ is $\frac12 \cdot 6 \cdot 8 = 24$ .
Find $AD$ . In a right triangle, the midpoint of the longest side is the center of the circle through all three corners. So $D$ is the same distance from $A$ , $B$ and $C$ : $AD = BD = CD = 5$ .
Areas of the halves. Triangles $ADB$ and $ADC$ have equal bases ( $BD = CD = 5$ ) and the same height from $A$ to line $BC$ , so each has area $24 \div 2 = 12$ .
Triangle $ADB$ has sides $6$ , $5$ , $5$ . Half the perimeter is $16 \div 2 = 8$ , so its radius is $12 \div 8 = \frac32$ .
Triangle $ADC$ has sides $8$ , $5$ , $5$ . Half the perimeter is $18 \div 2 = 9$ , so its radius is $12 \div 9 = \frac43$ .
Sum: $\frac32 + \frac43 = \frac96 + \frac86 = \frac{17}{6}$ .

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

Why this works

Whenever a problem asks for the radius of a circle drawn inside a triangle, reach for "area $=$ radius $\times$ half the perimeter." Two facts about right triangles give everything else: the midpoint of the longest side is equally far from all three corners, and a line from a corner to the middle of the opposite side cuts the area in half.

Alternative approach

Find each half's area directly. Triangle $ADB$ has two sides of $5$ and base $6$ ; the height to the base splits it into two $3$ - $4$ - $5$ right triangles, so the height is $4$ and the area is $\frac12 \cdot 6 \cdot 4 = 12$ . Triangle $ADC$ has two sides of $5$ and base $8$ ; the height splits it into $4$ - $3$ - $5$ right triangles, so the height is $3$ and the area is $\frac12 \cdot 8 \cdot 3 = 12$ . Then divide by half the perimeters as above.

The trap

Dividing the area by the whole perimeter instead of half of it, or assuming the two circles are the same size because the two halves have equal areas.

Concise 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 $\boxed{\textbf{(D)}\ \frac{17}{6}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.5)
Introduction to Geometry, Chapter 11: Circles (§11.3)

Techniques

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