Very Tricky Geometry Problem From Japan

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[marioalb]

Triangle APD is similar to triangle CPD, both with internal acute angles 45° and θ.

Triangle PCQ is an isósceles triangle, with internal acute angles θ and θ.

Then, angle AQD is 2θ
Right triangle AQD,
θ + 2θ = 90° = 3θ
θ = 30° ( Solved √ )

[antoniopedrofalcaolopesmor]

Assume square side measures 1.
Let PQ = QC = a.
Triangles ABP and QDP are similar and therefore we have
DQ/PQ = AB/PA
(1-a)/a = 1/PA
PA = a/(1-a)

AQ = AP+a = a/(1-a)+a = (a+a-a^2)/(1-a) =
= (2a-a^2)/(1-a) = a(2-a)/(1-a)

Let's use Pythagorean theorem for triangle ADQ :
1^2+(1-a)^2 = a^2*(2-a)^2/(1-a)^2
Simplifying and solving this equation for a we obtain a = 1-1/sqrt3.

Let angle theta = x
tgx= DQ/AD = (1-a)/1 = 1/sqrt3 = sqrt3/3
Thus, x = arctg(sqrt3/3) = 30 degrees.

[lasalleman]

Assume magnitude of side of the square = 1 Set up line PC as base of an isoceles triangle, BPC . Both sides = 1. Give you a base such that PC =.765 Now, each base angle = 67.5 degrees. Then, set up a new isoceles triangle with .765 (PC) as the base. You have 22.5 degrees at each end of the new triangle, 135 degrees at apex Q . Works out to PQ, and QC each = .414. So, QC, minus DC = Line DQ, which is = .586. Line DQ is sine of the angle needed. Line AD becomes the Cosine of the angle. Thus, going Pythagorean again. Sine/Cos = .586, resultant angle: 30.370. Call 30 degrees.

[quigonkenny]

As ∠DAB = 90° and ∠DAQ = θ, QAB = 90°-θ. As AB and DC are parallel, ∠AQD is an alternate interior angle to ∠QAB and thus ∠AQD = ∠QAB = 90°-θ. As ∠CDA = ∠ABC = 90° and BD, as a diagonal of sauare ABCD, bisects ∠CDA and ∠ABC, ∠ABD = ∠DBC = ∠CDB = ∠BDA = 90°/2 = 45°.

Draw CP. As AB = BC, ∠ABP = ∠PBC = 45°, and BP is common, ∆ABP and ∆PBC are congruent, and thus CP = PA and ∠BCP = ∠PAB = 90°-θ. As ∠BCD = 90°, ∠PCQ = θ.

As CQ = QP, ∆CQP is an isosceles teiangle and ∠QPC = ∠PCQ = θ. As ∠PQD is an exterior angle to ∆CQP at Q, ∠PQD = ∠QPC + ∠PCQ.

∠PQD = ∠QPC + ∠PCQ
90° - θ = θ + θ
90° = θ + θ + θ = 3θ
θ = 90°/3 = 30°

[Antony_V]

If we drop the perpendicular from P to the square's side AD and DC in points H and K we get two congruent triangles AHP and PKC. So angle PCK=θ as well as CPQ. Angle QPK=PAH =θ since AD and PK are parallel, thus 3θ=90°, θ=30°.

[RealQinnMalloryu]

360°ABCDQ/2 180°ABCDQ 3^60 3^20 3^2^10 3^2^2^5 3^1^2^1 32 (ABCDQ ➖ 3ABCDQ+2).

[SGuerra]

Que questão bonita. Parabéns pela escolha.

[michaeldoerr]

The angle theta is 30°. That is the first time in a while we had a congruence postulate and a problem with the angle is NOT 15°.

[giuseppemalaguti]

PQ=a..l=lato del quadrato risultato le due equazioni..ltgθ+a=l e (teorema del seno su PCB) la a, si semplifica l, risulta una equazione finale

[harikatragadda]

Draw the diagonal AC with O as its midpoint.
∆AOP is Congruent ∆COP by SAS.
Hence ∠PAO = ∠PCO, implying ∠DAP= ∠DCP=θ
Hence, ∠PQD =2θ= 90-θ
θ=30°

[professorrogeriocesar]

Ligue AC. Teorema do ângulo externo.
30.