Problem:
In △ABC, we have AB=1 and AC=2. Side BC and the median from A to BC have the same length. What is BC ?
Answer Choices:
A. 21+2
B. 21+3
C. 2
D. 23
E. 3 Solution:
Let M be the midpoint of BC, let AM=2a, and let θ=∠AMB. Then cos∠AMC=−cosθ. Applying the Law of Cosines to △ABM and to △AMC yields, respectively,
a2+4a2−4a2cosθ=1
and
a2+4a2+4a2cosθ=4
Adding, we obtain 10a2=5, so a=2/2 and BC=2a=2.
\mathrm
As above, let M be the midpoint of BC and AM=2a. Put a rectangular coordinate system in the plane of the triangle with the origin at M so that A has coordinates (0,2a). If the coordinates of B are (x,y), then the point C has coordinates (−x,−y),
So
x2+(2a−y)2=1 and x2+(2a+y)2=4
Combining the last two equations gives 2(x2+y2)+8a2=5. But, x2+y2=a2, so 10a2=5. Thus, a=2/2 and BC=2.
.jpg)