Difference between revisions of "1981 AHSME Problems/Problem 25"
Line 1: | Line 1: | ||
+ | __FORCETOC__ | ||
+ | |||
== Problem == | == Problem == | ||
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 |
Latest revision as of 21:01, 9 August 2023
Contents
[hide]Problem
In in the adjoining figure,
and
trisect
. The lengths of
,
and
are
,
, and
, respectively. The length of the shortest side of
is
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
.
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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 | ||
All AHSME Problems and Solutions |