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Difference between revisions of "1951 AHSME Problems/Problem 32"

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If <math>\triangle ABC</math> is inscribed in a semicircle whose diameter is <math>AB</math>, then <math>AC+BC</math> must be  
 
If <math>\triangle ABC</math> is inscribed in a semicircle whose diameter is <math>AB</math>, then <math>AC+BC</math> must be  
  
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<math> \textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} </math>      <math>
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<math> \textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} \qquad \textbf{(E)}\ AB^{2} </math>
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\textbf{(E)}\ AB^{2} </math>
 
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[[1951 AHSME Problems/Problem 32|Solution]]
 
  
 
==Solution==
 
==Solution==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:10, 21 June 2024

Problem

If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$, then $AC+BC$ must be

$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} \qquad \textbf{(E)}\ AB^{2}$

Solution

Because $AB$ is the diameter of the semi-circle, it follows that $\angle C = 90$. Now we can try to eliminate all the solutions except for one by giving counterexamples.

$\textbf{(A):}$ Set point $C$ anywhere on the perimeter of the semicircle except on $AB$. By triangle inequality, $AC+BC>AB$, so $\textbf{(A)}$ is wrong. $\textbf{(B):}$ Set point $C$ on the perimeter of the semicircle infinitesimally close to $AB$, and so $AC+BC$ almost equals $AB$, therefore $\textbf{(B)}$ is wrong. $\textbf{(C):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$, it follows that $\textbf{(C)}$ is wrong. $\textbf{(E):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$, it follows that $\textbf{(E)}$ is wrong. Therefore, the only possible case is $\boxed{\textbf{(D)}\ \leq AB\sqrt{2}}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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