A Geometry Problem from the
2007 Bay Area Mathematical Olympiad

Steven R. Dunbar
Department of Mathematics
Univesity of Nebraska-Lincoln
Lincoln, Nebraska
www.math.unl.edu/~sdunbar
sdunbar1 at unl dot edu

June 30, 2008

Problem: In △ABC, D and E are two points in the interior of side BC such that BD = CE and ⁄-BAD = ⁄CAE. Prove that △ABC is isosceles.

Solution: Without loss of generality, assume that the points B, D E and C are in that order on BC. (Otherwise exchange the labels on points D and E so the former E becomes D′ and the former D becomes E′. Then recognize that the common length ED can be subtracted from both lengths i.e BD - DE = BE = BD′ and CE - ED = CD = CE′. Likewise the common angle ⁄DAE can be subtracted from both angles to preserve the hypotheses.)

From vertex A to point F on the opposite side BC, let segment AF be the angle bisector of ⁄DAE. Let α = ⁄-BAD = ⁄CAE. Let β = ⁄DAF = ⁄EAF. I will use γ = ⁄AFC as a parameter and express several quantities in terms of this parameter and the fixed length AF.

Note that 0 < α < π∕2, 0 < β < π∕2 and α + β < π∕2. Note that if γ = α + β the line ←→ BC is parallel to the line ← → AB, an impossibility. Furthermore if γ < α + β, the intersection of line ←→ BC and line ←→ AB has “flipped” to the other side of A and now E and D are exterior to BC. Likewise π - γ > α + β or a similar argument applies to lines ←B→C and ←AC→. Summarizing: α + β < γ < π - (α + β).

Apply the Law of Sines to △F AC to obtain

F C AF ---------- = ----------. sin(F AC ) sin(ACF )

Therefore

sin(F AC ) F C = AF ---------- sin(ACF )

and likewise

FB = AFsin(F-AB-) sin(ABF )
FD = AFsin(F-AD-) sin(ADF )
FE = AFsin(F AE ) ---------- sin(AEF )

Then

CE = AF[sin(F AC ) sin(F AE )] ---------- - ---------- sin(ACF ) sin(AEF )
BD = AF[ ] sin(F AB ) sin(F AD ) sin(ABF--) - sin(ADF--).

Now using that α = ⁄BAD = ⁄CAE and since AF bisects ⁄-DAE, so ⁄FAD = ⁄FAE = β and ⁄FAB = ⁄-FAC = α + β. Let ⁄AFE = ⁄AFC = γ and ⁄AFB = ⁄AFD = π - γ. Then

CE = AF[ ] -----sin-(α +-β)------- -----sin(β)----- sin (π - (α + β + γ)) sin(π - (β + γ))
BD = AF[ sin (α + β) sin(β) ] -------------------------- - -------------------- sin (π - (π - γ) - (α + β)) sin(π - (π - γ) - β).

Simplifying

CE = AF[ ] --sin(α-+-β)--- --sin(β-)-- sin(α + β + γ) - sin(β + γ )
BD = AF[ ] ---sin(α-+-β)--- - ---sin(β)--- sin(- α - β + γ) sin(- β + γ).

Now equate these expressions since BD = CE

--sin-(α-+-β)---- --sin(β)-- = --sin(α-+--β)---- --sin(β)-- sin (α + β + γ) sin(β + γ) sin(γ - α - β) sin(γ - β)

or

--sin(α-+-β-)--- --sin(α-+-β)---= --sin(β-)--- --sin-(β-)--. sin(α + β + γ) sin (γ - α - β) sin(β + γ ) sin (γ - β)

Adding over a common denominator, and reducing trigonometric products to sums, obtain

4 cos(γ)sin2(α + β) 4cos(γ )sin2 (β ) ------------------------= ------------------ cos(2γ) - cos(2(α + β )) cos(2γ) - cos(2β)

Recalling that α + β < γ < π - (α + β), neither of the denominator expressions is 0.

The only way these two expressions can be equal is for cos(γ) = 0 or equivalently γ = π∕2, hence ⁄FCA = ⁄FBA and △ABC is isosceles.

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Commentary: Constructing this problem with GeoGebra showed me that the varying the base BC was the key to understanding the problem. That suggested rotating the base BC about the point F using the angle γ as a parameter to measure the rotation was a good way to organize the proof. The rest of the proof then organizes itself around clearly expressing the lengths BD and CE in terms of the parameter γ. Expressing the lengths yielded expressions such as sin(γ - α - β) in denominators. Again rotating segment BC about F shows that if γ = α + β the line BC is parallel to the line AB, so this parameter condition must be ruled out.