1999 AHSME Problems/Problem 26

Problem

Three non-overlaping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons for a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?

$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$

Solution

We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\circ}$ .

Let the number of sides in our polygons be $3\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$ .

The integral angle of a regular $k$ -gon is $180 \frac{k-2}k$ . Therefore we are looking for integer solutions to:

$360 = 180\left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)$

Which can be simplified to:

$2 = \left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)$

Furthermore, we know that two of the polygons are congruent, thus WLOG $a=c$ . Our equation now becomes

$2 = \left( 2\cdot\frac{a-2}a + \frac{b-2}b \right)$

Multiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$ .

Using the standard technique for Diophantine equations, we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$ .

Remembering that $a,b\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$ , $(2,4)$ , $(4,2)$ , and $(8,1)$ .

These correspond to the following pairs $(a,b)$ : $(5,10)$ , $(6,6)$ , $(8,4)$ , and $(12,3)$ .

The perimeters of the resulting polygon for these four cases are $14$ , $12$ , $14$ , and $\boxed{21}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1999 AHSME Problems/Problem 26

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