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Difference between revisions of "1961 AHSME Problems/Problem 38"

(Solution to Problem 38)
 
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== Problem 38==
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== Problem ==
  
 
<math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>.  
 
<math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>.  
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\textbf{(C)}\ s^2 \ge 8r^2 \qquad\\
 
\textbf{(C)}\ s^2 \ge 8r^2 \qquad\\
 
\textbf{(D)}\ s^2\le4r^2 \qquad
 
\textbf{(D)}\ s^2\le4r^2 \qquad
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\textbf{(E)}\ s^2=4r^2  </math>  
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\textbf{(E)}\ s^2=4r^2  </math>
  
 
==Solution==
 
==Solution==

Revision as of 23:27, 1 June 2018

Problem

$\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$. Point $C$ does not coincide with either $A$ or $B$. Let $s=AC+BC$. Then, for all permissible positions of $C$:

$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ \textbf{(D)}\ s^2\le4r^2 \qquad \textbf{(E)}\ s^2=4r^2$

Solution

[asy] draw((-50,0)--(-30,40)--(50,0)--(-50,0)); draw(Arc((0,0),50,0,180)); draw(rightanglemark((-50,0),(-30,40),(50,0),200)); dot((-50,0)); label("A",(-50,0),SW); dot((-30,40)); label("C",(-30,40),NW); dot((50,0)); label("B",(50,0),SE); [/asy] Since $s=AC+BC$, $s^2 = AC^2 + 2 \cdot AC \cdot BC + BC^2$. Since $\triangle ABC$ is inscribed and $AB$ is the diameter, $\triangle ABC$ is a right triangle, and by the Pythagorean Theorem, $AC^2 + BC^2 = AC^2 = (2r)^2$. Thus, $s^2 = 4r^2 + 2 \cdot AC \cdot BC$.


The area of $\triangle ABC$ is $\frac{AC \cdot BC}{2}$, so $2 \cdot [ABC] = AC \cdot BC$. That means $s^2 = 4r^2 + 4 \cdot [ABC]$. The area of $\triangle ABC$ can also be calculated by using base $AB$ and the altitude from $C$. The maximum possible value of the altitude is $r$, so the maximum area of $\triangle ABC$ is $r^2$.


Therefore, $s^2 \le 8r^2$, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions


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