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Difference between revisions of "1961 AHSME Problems/Problem 8"
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Let the two base angles of a triangle be ''A'' and ''B'', with ''B'' larger than ''A''. The altitude to the base divides the vertex angle ''C'' into two parts, <math>C_1</math> and <math>C_2</math>, with <math>C_2</math> adjacent to side ''a''. Then:
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== Problem ==
  
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(A) <math>C_1+C_2=A+B</math>
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Let the two base angles of a triangle be <math>A</math> and <math>B</math>, with <math>B</math> larger than <math>A</math>.
 +
The altitude to the base divides the vertex angle <math>C</math> into two parts, <math>C_1</math> and <math>C_2</math>, with <math>C_2</math> adjacent to side <math>a</math>. Then:
  
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(B) <math>C_1-C_2=B-A</math>
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<math>\textbf{(A)}\ C_1+C_2=A+B \qquad
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\textbf{(B)}\ C_1-C_2=B-A \qquad\\
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\textbf{(C)}\ C_1-C_2=A-B \qquad
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\textbf{(D)}\ C_1+C_2=B-A\qquad
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\textbf{(E)}\ C_1-C_2=A+B</math>
  
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(C) <math>C_1-C_2=A-B</math>
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==Solution==
  
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(D) <math>C_1+C_2=B-A</math>
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<asy>
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draw((0,0)--(120,0)--(40,50)--(0,0));
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draw((40,50)--(40,0));
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draw((40,4)--(44,4)--(44,0));
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label("$B$",(10,5));
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label("$A$",(100,5));
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label("$C_1$",(47,38));
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label("$C_2$",(34,33));
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</asy>
  
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(E) <math>C_1-C_2=A+B</math>
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Noting that side <math>a</math> is opposite of angle <math>A</math>, label the diagram as shown.
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From the two right triangles,
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<cmath>B + C_2 = 90</cmath>
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<cmath>A + C_1 = 90</cmath>
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Substitute to get
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<cmath>B + C_2 = A + C_1</cmath>
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<cmath>B - A = C_1 - C_2</cmath>
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The answer is <math>\boxed{\textbf{(B)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=7|num-a=9}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 13:36, 19 May 2018

Problem

Let the two base angles of a triangle be $A$ and $B$, with $B$ larger than $A$. The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$, with $C_2$ adjacent to side $a$. Then:

$\textbf{(A)}\ C_1+C_2=A+B \qquad \textbf{(B)}\ C_1-C_2=B-A \qquad\\ \textbf{(C)}\ C_1-C_2=A-B \qquad \textbf{(D)}\ C_1+C_2=B-A\qquad \textbf{(E)}\ C_1-C_2=A+B$

Solution

[asy] draw((0,0)--(120,0)--(40,50)--(0,0)); draw((40,50)--(40,0)); draw((40,4)--(44,4)--(44,0)); label("$B$",(10,5)); label("$A$",(100,5)); label("$C_1$",(47,38)); label("$C_2$",(34,33)); [/asy]

Noting that side $a$ is opposite of angle $A$, label the diagram as shown.

From the two right triangles, \[B + C_2 = 90\] \[A + C_1 = 90\] Substitute to get \[B + C_2 = A + C_1\] \[B - A = C_1 - C_2\] The answer is $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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