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Difference between revisions of "1959 AHSME Problems/Problem 40"

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== Problem ==
 
== Problem ==
  
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On the same side of a straight line three circles are drawn as follows: a circle with a radius of <math>4</math> inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: <math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12</math>
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In <math>\triangle ABC</math>, <math>BD</math> is a median. <math>CF</math> intersects <math>BD</math> at <math>E</math> so that <math>\overline{BE}=\overline{ED}</math>. Point <math>F</math> is on <math>AB</math>. Then, if <math>\overline{BF}=5</math>,  
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<math>\overline{BA}</math> equals:
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<math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}  </math>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:03, 21 July 2024

Problem

In $\triangle ABC$, $BD$ is a median. $CF$ intersects $BD$ at $E$ so that $\overline{BE}=\overline{ED}$. Point $F$ is on $AB$. Then, if $\overline{BF}=5$, $\overline{BA}$ equals: $\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these}$

Solution

Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.

Since ADG is similar to ACB, we know that AG is 10.

Thus, AF = 10 + 5 = 15, which is answer $\fbox{\textbf{(C)}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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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