Difference between revisions of "1983 AHSME Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
− | 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 | + | 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. |
==See Also== | ==See Also== |
Latest revision as of 15:20, 10 April 2021
Contents
Problem
How many non-congruent right triangles are there such that the perimeter in and the area in are numerically equal?
Solution 1
Let the triangle have legs of length and , so by the Pythagorean Theorem, the hypotenuse has length . Therefore we require Now, as and are side lengths of a triangle, they must both be non-zero, so we can safely divide by to give , so for any value of other than , we can generate a valid corresponding value of .
Notice also that each of these values of will give a unique corresponding value of , since , and by considering the graph of , 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 except ), so the answer is .
Solution 2
We use the formula , where is the area, is the inradius, and is the semiperimeter of a triangle. In this case, we have , so as and are nonzero (the triangle must not be degenerate), we must have . Now simply observe that there are clearly many right triangles that can be drawn with inradius (try drawing a circle of radius 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 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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