A Very Beautiful Geometry Problem

A Very Tricky Geometry Problem

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[quigonkenny]

Let O be the midpoint of DC. If we inscribe ∆ADC in a circle, since ∠CAD = 90°, DC is a diameter of circle O and OD = OC = OA = y/2.

Since OA = OC, ∆AOC is isosceles, and ∠CAO = ∠OCA = θ. Since ∠CAO = θ, ∠OAB = 2θ.

As ∠BOA is an exterior angle to ∆AOC at O, ∠BOA = ∠OCA + ∠CAO = θ + θ = 2θ.

As ∠BOA = ∠OAB = 2θ, ∆ABO is isosceles and AB = BO.

AB = BO
x + 5 = x + y/2
y/2 = 5
y = 2(5) = 10

[oguzhanbenli]

Hehe. X is zero, and this was all illustration and magic. 3 theta is 90 degrees, and theta is simply 30 degrees. That's why y is 10. X/sin 30. This is not a problem at all but spawning the rabbit out of the hat. If you want, you can take x as 1 billion, too. But don't forget to take the opposite side of angle theta as

[giuseppemalaguti]

queste 2 equazioni risulta Y=10.. indipendentemente da θ

[jonpress]

By the law of sines in ∆ABC, (x+5)/sin θ = (x+y)/sin 3θ = (x+y)/(3sin θ - 4sin^3 θ). Simplify a bit to get (x+y) / (x+5) = 3-4sin^2 θ.

By the law of sines in ∆ABD, (x+5)/sin (90+θ) = x/sin(3θ-90). Use various trig identities to convert this to (x+5)/cos θ = -x/cos 3θ, which leads to x / (x+5) = 3-4cos^2 θ.

Add these two results to get (2x+y)/(x+5) = 6 - 4(sin^2 θ + cos^2 θ) = 6 - 4 = 2. Solve (2x+y)/(x+5) = 2 to easily get y = 10.

Admittedly, using trig instead of raw geometry on difficult problems sometimes feels like cheating, but still.