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Difference between revisions of "1983 AHSME Problems/Problem 24"

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==Solution 2==
 
==Solution 2==
  
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We use the formula <math>A = rs</math>, where <math>A</math> is the area, <math>r</math> is the inradius, and <math>s</math> is the semiperimeter of a triangle. In this case, we have <math>A = 2s</math>, so as <math>A</math> and <math>s</math> are nonzero (the triangle must not be degenerate), we must have <math>r = 2</math>. Now simply observe that there are clearly many triangles that can be drawn with inradius <math>2</math> (try drawing a circle of radius <math>2</math> and then seeing how many triangles you can draw that have all three sides tangent to the circle at some point), so the answer must be <math>\boxed{\textbf{(E)} \ \text{infinitely many}}</math> as before.
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We use the formula <math>A = rs</math>, where <math>A</math> is the area, <math>r</math> is the inradius, and <math>s</math> is the semiperimeter of a triangle. In this case, we have <math>A = 2s</math>, so as <math>A</math> and <math>s</math> are nonzero (the triangle must not be degenerate), we must have <math>r = 2</math>. Now simply observe that there are clearly many right triangles that can be drawn with inradius <math>2</math> (try drawing a circle of radius <math>2</math> tangent to two lines that form a right angle and then seeing how many lines tangent to the circle you can draw to form the third side of the triangle), so the answer must be <math>\boxed{\textbf{(E)} \ \text{infinitely many}}</math> as before.
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==See Also==
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{{AHSME box|year=1983|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 14:20, 10 April 2021

Problem

How many non-congruent right triangles are there such that the perimeter in $\text{cm}$ and the area in $\text{cm}^2$ are numerically equal?

$\textbf{(A)} \ \text{none} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ \text{infinitely many}$

Solution 1

Let the triangle have legs of length $a$ and $b$, so by the Pythagorean Theorem, the hypotenuse has length $\sqrt{a^2+b^2}$. Therefore we require \begin{align*} &a + b + \sqrt{a^2+b^2} = \frac{1}{2} ab \\ \Rightarrow \quad &2 \sqrt{a^2+b^2} = ab - 2a - 2b \\ \Rightarrow \quad &4a^2 + 4b^2 = a^{2}b^{2} + 4a^2 + 4b^2 - 4a^{2}b - 4ab^{2} + 8ab \\ \Rightarrow \quad &4a^{2}b + 4ab^{2} = a^{2}b^{2} + 8ab. \end{align*} Now, as $a$ and $b$ are side lengths of a triangle, they must both be non-zero, so we can safely divide by $ab$ to give $4a + 4b = ab + 8 \Rightarrow b(a-4) = 4a-8 \Rightarrow b = \frac{4a-8}{a-4}$, so for any value of $a$ other than $4$, we can generate a valid corresponding value of $b$.

Notice also that each of these values of $a$ will give a unique corresponding value of $b$, since $\frac{4a-8}{a-4} = 4 + \frac{8}{a-4}$, and by considering the graph of $y = 4 + \frac{8}{x-4}$, it is clear that any horizontal line will intersect it at most once. Thus there are infinitely many valid solutions (one for every value of $a$ except $4$), so the answer is $\boxed{\textbf{(E)} \ \text{infinitely many}}$.

Solution 2

We use the formula $A = rs$, where $A$ is the area, $r$ is the inradius, and $s$ is the semiperimeter of a triangle. In this case, we have $A = 2s$, so as $A$ and $s$ are nonzero (the triangle must not be degenerate), we must have $r = 2$. Now simply observe that there are clearly many right triangles that can be drawn with inradius $2$ (try drawing a circle of radius $2$ tangent to two lines that form a right angle and then seeing how many lines tangent to the circle you can draw to form the third side of the triangle), so the answer must be $\boxed{\textbf{(E)} \ \text{infinitely many}}$ as before.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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