Triangles with Equal Area
The applet below illustrates a Problem 4 from the 2007 International Mathematics Olympiad:
In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area. |
What if applet does not run? |
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Copyright © 1996-2017 Alexander BogomolnySolution
The solution that I came up with is purely trigonometric, although I can't escape a feeling there must be a more elegant synthetic solution that employs some kind of shearing. (There is a solution that eschews trigonometry and equally simple another one. Finally, there is a another solution nearest to my expectations.)
What if applet does not run? |
Let O be the circumcenter of ΔABC. ΔCOR is isosceles, with CO = OR = R, the circumradius of ΔABC. Denote the angles of the latter α, β, γ. The angular measure of arc AR is γ, that of AC and BC 2β and 2α, respectively. The central angle COR is either
CR = 2R sin(α + γ/2) = 2R sin(α + γ/2). |
(The latter identity holds because
Setting, as usual, BC = a and AC = b, we get from triangles CKP and CLQ
CP = CK/cos(γ/2) = a / 2cos(γ/2), CQ = CL/cos(γ/2) = b / 2cos(γ/2). |
Now, let's turn to the triangles in question - RPK and RQL. We shall computer their areas through the standard formula,
PR = CR - CP, QR = CR - CQ. |
Further,
2 Area( ΔRPK) | = PR × h_{a} | |
= (2R sin(α + γ/2) - a / 2cos(γ/2)) × a·sin(γ/2)/2. |
Trying to simplify leads to
8 Area( ΔRPK)×cos(γ/2) / sin(γ/2) | = 4Ra sin(α + γ/2) cos(γ/2) - a² | ||
= 2Ra (sin(α + γ) + sin(α)) - a² | |||
= 2Ra (sin(β) + sin(α)) - a². |
By the Law of Sines,
8 Area( ΔRPK)×cos(γ/2) / sin(γ/2) | = 2Ra (sin(β) + sin(α)) - a² | ||
= ab + a² - a² | |||
= ab. |
So that Area( ΔRPK) = ab tan(γ/2) / 8. Obviously the same holds for Area( ΔRQL). Therefore, the two are indeed equal.
2007 IMO, Problem 4
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