1981 AHSME Problems/Problem 25
- 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)