Difference between revisions of "1981 AHSME Problems/Problem 25"
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− | + | == Problem 25 == | |
In <math>\triangle ABC</math> in the adjoining figure, <math>AD</math> and <math>AE</math> trisect <math>\angle BAC</math>. The lengths of <math>BD</math>, <math>DE</math> and <math>EC</math> are <math>2</math>, <math>3</math>, and <math>6</math>, respectively. The length of the shortest side of <math>\triangle ABC</math> is | In <math>\triangle ABC</math> in the adjoining figure, <math>AD</math> and <math>AE</math> trisect <math>\angle BAC</math>. The lengths of <math>BD</math>, <math>DE</math> and <math>EC</math> are <math>2</math>, <math>3</math>, and <math>6</math>, respectively. The length of the shortest side of <math>\triangle ABC</math> is | ||
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<math>\textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}</math> | <math>\textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}</math> | ||
− | + | == Solution == | |
Let <math>AC=b</math>, <math>AB=c</math>, <math>AD=d</math>, and <math>AE=e</math>. Then, by the Angle Bisector Theorem, <math>\frac{c}{e}=\frac{2}{3}</math> and <math>\frac{d}{b}=\frac12</math>, thus <math>e=\frac{3c}2</math> and <math>d=\frac b2</math>. | Let <math>AC=b</math>, <math>AB=c</math>, <math>AD=d</math>, and <math>AE=e</math>. Then, by the Angle Bisector Theorem, <math>\frac{c}{e}=\frac{2}{3}</math> and <math>\frac{d}{b}=\frac12</math>, thus <math>e=\frac{3c}2</math> and <math>d=\frac b2</math>. |
Revision as of 01:49, 9 August 2023
Problem 25
In in the adjoining figure,
and
trisect
. The lengths of
,
and
are
,
, and
, respectively. The length of the shortest side of
is
[asy] defaultpen(linewidth(.8pt)); pair A = (0,11); pair B = (2,0); pair D = (4,0); pair E = (7,0); pair C = (13,0); label("",A,N); label("
",B,SW); label("
",C,SE); label("
",D,S); label("
",E,S); label("
",midpoint(B--D),N); label("
",midpoint(D--E),NW); label("
",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E); [/asy]
Solution
Let ,
,
, and
. Then, by the Angle Bisector Theorem,
and
, thus
and
.
Also, by Stewart’s Theorem, and
. Therefore, we have the following system of equations using our substitution from earlier:
.
Thus, we have:
.
Therefore, , so
, thus our first equation from earlier gives
, so
, thus
. So,
and the answer to the original problem is
.
Aops-g5-gethsemanea2 (talk) 01:47, 9 August 2023 (EDT)