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/ sdunbar1
sdunbar1 at unl dot edu

January 27, 2009

Problem: This is problem 3 on the 2007 Bay Area Mathematical Olympiad, located at  http://www.bamo.org/attachments/bamo2007examsol.pdf  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.

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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 < π2, 0 < β < π2 and α + β < π2. 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¯. Thus γ > α + β. Likewise π γ > α + β or a similar argument applies to lines BC and AC. Summarizing: α + β < γ < π (α + β).

Apply the Law of Sines to FAC to obtain

FC sin(FAC) = AF sin(ACF).


FC = AF sin(FAC) sin(ACF)

and likewise

FB = AF sin(FAB) sin(ABF) FD = AF sin(FAD) sin(ADF) FE = AF sin(FAE) sin(AEF).


CE = AF sin(FAC) sin(ACF) sin(FAE) sin(AEF) BD = AF sin(FAB) sin(ABF) sin(FAD) 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(π (α + β + γ)) sin(β) sin(π (β + γ)) BD = AF sin(α + β) sin(π (π γ) (α + β)) sin(β) sin(π (π γ) β) .


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

Now equate these expressions since BD = CE

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


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

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

4 cos(γ) sin 2(α + β) cos(2γ) cos(2(α + β)) = 4 cos(γ) sin 2(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 = π2, hence FCA = FBA and ABC is isosceles.

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.

The official source at  http://www.bamo.org/attachments/bamo2007examsol.pdf  gives 4 distinct solutions, each shorter than the proof above. The solution above is a combination of their Solution 2 and Solution 3. However, each of their solutions gives a proof by contradiction, while the proof above is essentially direct. The proof above is dynamic in nature, based on the dynamic geometry view of the problem from GeoGebra by rotating the base. This indicates how the technology used to view the problem can influence the thinking and the derived proof.