2021 AMC 10B Problem 20

P16-20~6 minPrint

Quadrilaterals & Polygon Areas Angles & Polygons Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Cut the figure into known shapes (triangles, rectangles, sectors)

The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$

Show solution without answering

Key insight

Triangles ABF and BCF are equilateral, so the diagonal AC is 2√3, and the same holds for AD. Cut the pentagon into triangles ABC, ACD, ADE.

Solution

Step 1: find the equilateral triangles.

Name the two inside points: $F$ is joined to $A$ , $B$ and $C$ , and $G$ is joined to $A$ , $E$ and $D$ . The $11$ segments are the $5$ sides of the pentagon and these $6$ inside segments, all of length $2$ .
Triangle $ABF$ has sides $AB$ , $BF$ , $AF$ , all equal to $2$ , so it is equilateral. Triangle $BCF$ has sides $BC$ , $CF$ , $BF$ , all $2$ , so it is equilateral too.
The same happens on the other side: triangles $AEG$ and $DEG$ are equilateral.

Step 2: find the diagonal $AC$ .

Put the two equilateral triangles $ABF$ and $BCF$ together along $BF$ . They form a diamond shape $ABCF$ with all sides $2$ .
Its long diagonal $AC$ crosses the shared side $BF$ at a right angle and is made of the two heights of the equilateral triangles. The height of an equilateral triangle with side $2$ is $\sqrt{2^2 - 1^2} = \sqrt3$ . So $AC = 2\sqrt3$ .
By the same reasoning, $AD = 2\sqrt3$ .

Step 3: cut the pentagon into three triangles from $A$ .

Triangle $ABC$ . The diagonal $AC$ cuts the diamond $ABCF$ exactly in half. The diamond is two equilateral triangles, each of area $\frac12 \cdot 2 \cdot \sqrt3 = \sqrt3$ , so the diamond has area $2\sqrt3$ and triangle $ABC$ has area $\sqrt3$ .
Triangle $ADE$ . The same way, its area is $\sqrt3$ .
Triangle $ACD$ . Its sides are $AC = AD = 2\sqrt3$ and $CD = 2$ . The height from $A$ meets $CD$ at its midpoint, so by the Pythagorean theorem the height is $\sqrt{(2\sqrt3)^2 - 1^2} = \sqrt{12 - 1} = \sqrt{11}$ . The area is $\frac12 \cdot 2 \cdot \sqrt{11} = \sqrt{11}$ .

Step 4: add up.

The total area is $\sqrt3 + \sqrt3 + \sqrt{11} = 2\sqrt3 + \sqrt{11}$ .
The answer must look like $\sqrt m + \sqrt n$ , so write $2\sqrt3 = \sqrt{4 \cdot 3} = \sqrt{12}$ . The area is $\sqrt{12} + \sqrt{11}$ , and $m + n = 12 + 11 = 23$ .

The answer is $\boxed{\textbf{(D)}\ 23}$ .

Why this works

When many segments have the same length, look for equilateral triangles first. They give you angles and lengths, such as the diagonal $2\sqrt3$ . With the diagonals from $A$ known, the pentagon splits into three triangles you can handle with basic tools. The form $\sqrt m + \sqrt n$ is a hint to move the $2$ inside the square root.

Concise solution

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

Let $F$ be the interior point joined to $A$ , $B$ , $C$ and $G$ the interior point joined to $A$ , $E$ , $D$ . The eleven segments are the five sides plus $AF, BF, CF, AG, EG, DG$ , all of length $2$ .

Triangles $ABF$ and $BCF$ have all sides equal to $2$ , so both are equilateral. Therefore $\angle ABC = 60^\circ + 60^\circ = 120^\circ$ , and $ABCF$ is a rhombus with side $2$ and a $60^\circ$ angle at $A$ ; its long diagonal is $AC = 2\cdot 2\cos 30^\circ = 2\sqrt{3}$ . By the same reasoning on the right, $\angle AED = 120^\circ$ and $AD = 2\sqrt{3}$ .

Now split the pentagon by the diagonals from $A$ :

$[ABC] = \frac12 \cdot 2 \cdot 2 \sin 120^\circ = \sqrt{3}$ , and likewise $[ADE] = \sqrt{3}$ .
Triangle $ACD$ is isosceles with $AC = AD = 2\sqrt{3}$ and base $CD = 2$ . Its altitude from $A$ is $\sqrt{12 - 1} = \sqrt{11}$ , so $[ACD] = \frac12 \cdot 2 \cdot \sqrt{11} = \sqrt{11}$ .

Total area: $2\sqrt{3} + \sqrt{11} = \sqrt{12} + \sqrt{11}$ , so $m + n = 12 + 11 = 23$ .

The answer is $\boxed{\textbf{(D)}\ 23}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)