1959 AHSME Problems/Problem 10

Problem

In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$ , a point $D$ is taken on $AB$ at a distance $1.2$ from $A$ . Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$ . Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$

Solution

$[asy] import geometry; point B = (0,0); point C = (10,0); point A = (5,12); point D = 2*A/3; point E = 3(C-A)+A; // Triangle ABC draw(A--B--C--A); dot(A); label("A",A,N); dot(B); label("B",B,SW); dot(C); label("C",C,SE); // Triangle ADE draw(A--D--E--A); dot(D); label("D",D,NW); dot(E); label("E",E,SE); [/asy]$

Note that $[\triangle ABC]=[\triangle ADE]$ , so $\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE$ . Since $\angle BAC = \angle DAE$ , we have $AB*AC = AD*AE$ , so that $3.6*3.6 = 1.2*AE$ . Therefore, $AE = \frac{3.6^2}{1.2}=\boxed{\textbf{(D) }10.8}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1959 AHSME Problems/Problem 10

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