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Difference between revisions of "1999 AHSME Problems/Problem 26"

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==Problem==
 
==Problem==
  
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Three non-overlaping regular plane polygons, at least two of which are congruent, all have sides of length <math>1</math>. The polygons meet at a point <math>A</math> in such a way that the sum of the three interior angles at <math>A</math> is <math>360^{\circ}</math>. Thus the three polygons for a new polygon with <math>A</math> as an interior point. What is the largest possible perimeter that this polygon can have?  
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Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length <math>1</math>. The polygons meet at a point <math>A</math> in such a way that the sum of the three interior angles at <math>A</math> is <math>360^{\circ}</math>. Thus the three polygons form a new polygon with <math>A</math> as an interior point. What is the largest possible perimeter that this polygon can have?  
  
 
<math> \mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24</math>
 
<math> \mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24</math>

Latest revision as of 09:19, 23 January 2024

Problem

Three non-overlapping 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 form 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 $\boxed{21}$.

Solution 2

We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\boxed{21}$ sides.

See Also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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