2017 AMC 10B Problem 21

P21-25~5 minPrint

Triangle Centers & Cevians Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed Cut the figure into known shapes (triangles, rectangles, sectors)

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 $\triangle ADC$ ?

Show solution without answering

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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)