2007 AMC 10B Problem 7

P1-10~3 minPrint

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

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$

Show solution without answering

Key insight

Draw the line from C to E. It cuts the pentagon into a square and an equilateral triangle, so the angle at E is 90 plus 60.

Solution

All five sides are the same length; call it $s$ . Two right angles next to each other suggest a square, so look for one.

1. Draw the segment from $C$ to $E$ . It splits the pentagon into a four-sided piece $ABCE$ and a triangle $CDE$ .
2. In $ABCE$ , the angles at $A$ and $B$ are both $90^\circ$ . So side $EA$ and side $BC$ both stand straight up from $AB$ , on the same side.
3. Those two sides are both $s$ long. So $E$ and $C$ are the same height above $AB$ , and $EC$ is parallel to $AB$ and also $s$ long. That makes $ABCE$ a square with side $s$ .
4. So $CE = s$ , and the square's corner at $E$ is $\angle AEC = 90^\circ$ .
5. Now triangle $CDE$ has $CD = s$ , $DE = s$ and $CE = s$ . All three sides are equal, so it is equilateral and $\angle CED = 60^\circ$ .
6. The pentagon's angle at $E$ is made of the two pieces:
$\angle E = 90^\circ + 60^\circ = 150^\circ.$

The answer is $\boxed{\textbf{(E)}\ 150}$ .

Why this works

Equal sides plus a pair of right angles is a strong hint that a square is hidden inside. One well-chosen diagonal shows it and leaves a triangle you also know completely. The angle sum of a pentagon alone would only tell you that the angles at $C$ , $D$ and $E$ add to $360^\circ$ , which is not enough.

Concise solution

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

Let every side have length $s$ and draw the diagonal $CE$ .

Look at quadrilateral $ABCE$ . It has $EA = AB = BC = s$ and right angles at $A$ and $B$ . Two right angles at the ends of $AB$ force $EA \parallel BC$ , and since those two sides are also equal, $ABCE$ is a parallelogram with a right angle and two adjacent sides equal, that is, a square. So $CE = s$ and $\angle AEC = 90^\circ$ .

Now triangle $CDE$ has $CD = DE = CE = s$ : it is equilateral, so $\angle CED = 60^\circ$ .

The pentagon's angle at $E$ is made of these two pieces:
$\angle E = \angle AEC + \angle CED = 90^\circ + 60^\circ = 150^\circ.$

The answer is $\boxed{\textbf{(E)}\ 150}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 9: Polygons (§9.1, §9.2)