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)