Difference between revisions of "1959 AHSME Problems/Problem 10"
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| + | == Problem == | ||
| + | |||
In <math>\triangle ABC</math> with <math>\overline{AB}=\overline{AC}=3.6</math>, a point <math>D</math> is taken on <math>AB</math> at a distance <math>1.2</math> from <math>A</math>. Point <math>D</math> is joined to <math>E</math> in the prolongation of <math>AC</math> so that <math>\triangle AED</math> is equal in area to <math>ABC</math>. Then <math>\overline{AE}</math> is: <math>\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6</math> | In <math>\triangle ABC</math> with <math>\overline{AB}=\overline{AC}=3.6</math>, a point <math>D</math> is taken on <math>AB</math> at a distance <math>1.2</math> from <math>A</math>. Point <math>D</math> is joined to <math>E</math> in the prolongation of <math>AC</math> so that <math>\triangle AED</math> is equal in area to <math>ABC</math>. Then <math>\overline{AE}</math> is: <math>\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6</math> | ||
== 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} = | + | <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 <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>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1959|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 12:20, 21 July 2024
Problem
In 
with 
, a point 
is taken on 
at a distance 
from 
. Point is joined to 
in the prolongation of 
so that 
is equal in area to 
. Then 
is: 
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]](http://latex.artofproblemsolving.com/6/3/6/636d1f23a4a183d1e0511c4dcdf069f8cb9d7cc3.png)
Note that ![$[\triangle ABC]=[\triangle ADE]$](http://latex.artofproblemsolving.com/8/7/1/871c9f7bd8d03ab287154e9f76cbb59f3638ea9c.png)
, so 
. Since 
, we have 
, so that 
. Therefore, 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
