Difference between revisions of "1959 AHSME Problems/Problem 10"

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== Solution ==  
 
== Solution ==  
  
Note that <math>\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE</math>. Since <math>\angle BAC = \angle DAE</math>, we have <math>AB*AC = AD*AE</math>, so that <math>3.6*3.6 = 1.2*AE</math>. Therefore, <math>AE = \frac{3.6^2}{1.2} = 10.8</math>. Thusly, our answer is <math>\boxed{\text{(D)}}</math>, and we are done.
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<asy>
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import geometry;
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point B = (0,0);
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point C = (10,0);
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point A = (5,12);
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point D = 2*A/3;
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point E = 3(C-A)+A;
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// Triangle ABC
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draw(A--B--C--A);
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dot(A);
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label("A",A,N);
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dot(B);
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label("B",B,SW);
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dot(C);
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label("C",C,SE);
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// Triangle ADE
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draw(A--D--E--A);
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dot(D);
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label("D",D,NW);
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dot(E);
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label("E",E,SE);
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</asy>
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Note that <math>[\triangle ABC]=[\triangle ADE]</math>, so <math>\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE</math>. Since <math>\angle BAC = \angle DAE</math>, we have <math>AB*AC = AD*AE</math>, so that <math>3.6*3.6 = 1.2*AE</math>. Therefore, <math>AE = \frac{3.6^2}{1.2}=\boxed{\textbf{(D) }10.8}</math>.
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==See also==
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{{AHSME 50p box|year=1959|num-b=9|num-a=11}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 12:20, 21 July 2024

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, $AE = \frac{3.6^2}{1.2}=\boxed{\textbf{(D) }10.8}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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