Japanese Math Olympiad Challenge | A Very Nice Geometry Problem

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Пікірлер: 23

@imetroangola17 2 ай бұрын

Solution: Note that ∠APB= 2θ. ∆ABP is rectangular, therefore: PB = 8/sin 2θ, on the other hand, in the rectangle ∆AQC, we have: QB = 5/cos θ. Thus, in the right triangle ∆BPQ: cos θ = QB/PB cos θ = 5/cos θ ÷ 8/sin 2θ cos θ = (5/cos θ) × (2sin θ cos θ/8) cos θ = 5 sin θ/4 sin θ/cos θ = 4/5→ tg θ = 4/5. In the right triangle ∆BQC: tg θ = QC/BC → 4/5 = QC/5 QC = 4 → DQ = 4. In the right triangle ∆PDQ, we have: tg θ = PD/DQ → 4/5 = PD/4 PD = 16/5. Therefore, [PDQ] = PD × DQ/2 = 16/5 × 4/2 *[PDQ] = 32/5 square units*

@ناصريناصر-س4ب 2 ай бұрын

Suppose QC=x and PD=y, from which QD=8-x. Triangles BQC and PDQ are similar, from which 5/x=(8-x)/y, so 5y=8x-x². Line BC intersects line PQ at point S. Triangles PDQ and CQS are congruent, from which QC=QD, from which 8-x=x, so x=4, from which y=16/5, so the area of triangle PDQ is equal to ((16/5)*4)/2=32/5.

@zehradiyab3439 2 ай бұрын

I proof DC=4 in another way If we put PB is a diameter of circle which have two right inscribed angles, If we put O is the centre of the circle which passes through the points A,B,P andQ, So AO and BO are radii of the circle, so ABO is isosceles triangle. If we take the point E as a midpoint of AB, so OE is perpendicular to AB, and OE is parallel to BC ...1 OB=OQ=r so angle OBQ=angle OQB , so angle CBQ=angle OQB, so CQ is parallel toBC ...2 From 1 and 2 EQ which passes through the point O is parallel to BC, so angle EQC=90° and EQ is perpendicular to DC, so AEQD is rectangle, and DQ=AE=½AB=½×8=4

@murdock5537 2 ай бұрын

Nice one!

@marcgriselhubert3915 2 ай бұрын

Let's use an orthonormal center B, first axis (BA), second axis (BC), and let's note t = tan(theta). We have QC = BC.t = 5.t in triangle BCQ We also have AP = AB.tan(90° -2.theta) = 8.cotan(2.theta) = 8.(1/tan(2.theta)) = 8.((1 -t^2)/2.t) = (4 -4.t^2)/t = (4/t) - 4.t We have Q(5.t; 5), P(8; (4/t) -4.t), VectorBQ(5.t; 5), VectorQP(8 -5.t; (4/t) - 4.t -5) These vectors are orthogonal, so t(8 -5.t) + ((4/t) -4.t -5) = 0 or 5.(t^3) -4.(t^2) +5.t -4 = 0 or (t^2 + 1).(5.t - 4) = 0, which gives that t = 4/5. Then AP = 5 - 16/5 = 9/5 and DP = 5 - 9/5 = 16/5, and CQ = 4 and DQ = 8 - 4 = 4. The area of DQP is (1/2).DQ.DP = (1/2).4.(16/5) = 32/5.

@mochemiguel1233 2 ай бұрын

Extend PQ, and BC and intersect in M point, and if you notice, BPQ and BMQ are congruents, CM = a (it'll be used later), and PDQ and QCM are congruents too, so DQ = QC = 4 (You got half of problem). Then, BM = BP = 5+a (a is CM), AP= 5-a and you can use PIT, (5+a)2 = (5-a)2 + (8)2 ==> a = 3.2. Then Area is 4 x 3.2 / 2 == 32/5

@2012tulio Ай бұрын

Let PD = x , DQ= y Tan theta= x/y ----eq1 Tan theta= 8-y / 5 -----eq2 Tan 2theta= 8/ 5-x -----eq3 By solving these 3 equations together we get x = 3.2 and y = 4.2 So the area of triangle PDQ is 0.5 * 3.2* 4.2 = 6.7

@marioalb9726 2 ай бұрын

Angles θ only can be equals, if position of point Q, is equidistant in the segment DC. Only possibility !!! For other ratio 5/8, angles θ can't be equals. b = 8/2 = 4 cm h = 4*4/5= 16/5 cm A = ½b.h = ½*4*16/5 A = 32/5 cm² ( Solved √ )

@oscarcastaneda5310 2 ай бұрын

I determined that tan(theta) = 4/5. From this , PD = 16/5 and DQ = 4. The Area calculates to 32/5.

@ludmilaivanova1603 2 ай бұрын

I found x=4 by just using similarrity of triangles BCQ, PDQ and BQP and also using DQ=x and QC= 8-x. No additional constructions.

@himo3485 Ай бұрын

BEQ≅BCQ 　　　 BC=BE=5 PEQ≅PDQ CQ=EQ=DQ=x x+x=8 2x=8 x=4　　　　　 QDP∞BCQ 4/5 = PD/4 PD=16/5 area of triangle DPQ = 4*16/5*1/2 = 32/5

@giuseppemalaguti435 2 ай бұрын

Solita procedura trigonometrica...Ablue=differenza di 40-(somma dei 3 triangoli)=40-32(ctg2θ)^2-(25/2)tgθ-10(secθ)^2...con tgθ=4/5...Ablue=40-36/5-10-82/5=(200-36-50-82)/5=32/5...come sempre,credo ci saranno metodi piu veloci

@FFernandezB 2 ай бұрын

Three similar triangles, write three proportions and solve the equation

@johnoehrle5973 Ай бұрын

PQ=BQ*QC/5. DQ=PQ*5/BQ=QC=4. DP=4/5*4...

@michaeldoerr5810 2 ай бұрын

Hello, I got the answer before you did. The area was 32/5 units square. And it looks like HL congruence has to be used to calculate x. And HL similarity is used to compute the other side. A pair of congruent triangles was required. I hope that this makes sense.