Problem:
Triangle ABC has ∠C=60∘ and BC=4. Point D is the midpoint of BC. What is the largest possible value of tan(∠BAD)?
Answer Choices:
A. 63
B. 33
C. 223
D. 42−33
E. 1 Solution:
Let C=(0,0),B=(2,23), and A=(x,0) with x>0. Then D=(1,3). Let P be on the positive x-axis to the right of A. Then ∠BAD=∠PAD−∠PAB. Provided ∠PAD and ∠PAB are not right angles, it follows that
Therefore the largest possible value of tan(∠BAD) is 3/(42−3).
\section*{OR}
Because the circle with diameter BD does not intersect the line AC, it follows that ∠BAD<90∘. Thus the value of tan(∠BAD) is greatest when ∠BAD is greatest. This occurs when A is placed to minimize the size of the circle passing through A,B, and D, so the maximum is attained when that circle is tangent to AC at A. For this location of A, the Power of a Point Theorem implies that
AC2=CB⋅CD=4⋅2=8, and AC=8=22
Because CBCA=CACD, it follows that △CAD is similar to △CBA. Thus AB=2AD. The Law of Cosines, applied to △ADC, gives
AD2=CD2+CA2−2CD⋅CA⋅cos60∘=12−42
Let O be the center of the circle passing through A,B, and D. The Extended Law of Sines, applied to △ABD and △ADC, gives