## 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.

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Solution

### Solution

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.)

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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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 2β + γ or 2α + γ. The base CR of ΔCOR is then found to be

 CR = 2R sin(α + γ/2) = 2R sin(α + γ/2).

(The latter identity holds because (α + γ/2) + (β + γ/2) = 180° and sin(x) = sin(180° - x).)

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, Area = Base × Altitude / 2. Take for the base the sides PR and QR and observe that the altitudes equal ha = a·sin(γ/2)/2 and hb = b·sin(γ/2)/2, respectively while

 PR = CR - CP, QR = CR - CQ.

Further,

 2 Area( ΔRPK) = PR × ha = (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, 2R sin(α) = a and 2R sin(β) = b. Thus we may continue

 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.

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