Problem:
In β³ABC, 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+2ββ
B. 21+3ββ
C. 2β
D. 23β
E. 3β Solution:
Let M be the midpoint of BC, let AM=2a, and let ΞΈ=β AMB. Then cosβ AMC=βcosΞΈ. Applying the Law of Cosines to β³ABM and to β³AMC yields, respectively,
a2+4a2β4a2cosΞΈ=1
and
a2+4a2+4a2cosΞΈ=4
Adding, we obtain 10a2=5, so a=2β/2 and BC=2a=2ββ.
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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 (βx,βy),
So
x2+(2aβy)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=2β/2 and BC=2ββ.
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