Spain Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods

[@RobertHering-tq7bn]

This is another possible but more complex way to solve the problem. The main idea is to see all the points coordinates in x and y. Here are the main points... O (0,0) - P (0,1) - B (2,0) - Q (Qx,Qy) and the main question is to find Qx and Qy. Based on the fact that angle PQB is 90°, we have the following relations between x and y: (1) x² + y² = 4 (2) straight line through P and Q: (y-1)/(x-0) = m y = mx + 1 (3) straight line through Q and B: (y-0)/(x-2) = -(1/m) y = -(1/m)x + 2/m (2)/(3): mx+1 = -(1/m)x + 2/m x = (2-m)/(1+m²) This gives with (2) for y: y = (2m-m²)/(1+m²) + 1 = (2m - m² + 1 + m²)/(1+m²) = (2m+1)/(1+m²) Using this in (1) gives... (2-m)²/(1+m²)² + (2m+1)²/(1+m²)² = 4 (5+5m²)/(1+m²)² = 4 5(1+m²)/(1+m²)² = 4 5/(1+m²) = 4 This gives us finally... m = 1/2 And we get with m used in (2) and (3)... (4) y = (1/2)x + 1 (5) y = -2x + 4 In the end we get for Qx and Qy... (1/2)Qx+1 = -2Qx+4 (5/2)Qx =3 Qx = 6/5 and Qy = (1/2)(6/5)+1 = 8/5 So, PQ² = (8/5 - 1)² + (6/5 - 0)² = (3/5)² + (6/5)² = 45/25 = 9/5 and PQ = 3/sqrt(5) and QB² = (0 - 8/5)² + (2 - 6/5)² = 64/25 + 16/25 = 80/25 = 16/5 and QB = 4/sqrt(5) F(PQB) = (1/2)*(3/sqrt(5))*(4/sqrt(5) = 6/5

[@prossvay8744]

Draw a circle In right triangle OBP So BP=√5 In right triangle OCP So CP=√5 Let QP=a a√5=1(3) So a=3/√5 So BQ=√(√5)^2-(3/√5)^2=4√5/5 So triangle area=1/2(3√5/5)(4√5/5)=6/5=1.2❤❤❤

[@quigonkenny]

Extend quarter circle O to a semicircle, with M as the point opposite B on the diameter. As Q is a point on the circumference and ∠BQP = 90°, then by Thales' Theorem, if QP is extended, it will intersect with point M, as BM is a diameter. As OM = OB = 2, ∠POB = ∠MOP = 90°, and OP is common, ∆POB and ∆MOP are fongruent. As ∠OPB = ∠QMO and ∠POB = ∠BQM = 90°, ∆BQM and ∆POB are similar triangles. Triangle ∆POB: OP² + OB² = BP² 1² + 2² = BP² BP² = 1 + 4 = 5 BP = √5 BQ/MB = OP/BP BQ/4 = 1/√5 BQ = 4/√5 QM/BQ = OB/OP QM/(4/√5) = 2/1 = 2 QM = 2(4/√5) = 8/√5 The area of ∆BQP is equal to the area of ∆BQM minus the areas of congruent triangles ∆POB and ∆MOP. Triangle ∆BQP: A = QM(BQ)/2 - OB(OP)/2 - OM(OP)/2 A = (8/√5)(4/√5)/2 - 2(1)/2 - 2(1)/2 A = 16/5 - 1 - 1 = (16-10)/5 = 6/5 sq units

[@richardleveson6467]

I always enjoy your presentations here - your constructions are ingenious and the explanations always clear. Thank you!

[@santiagoarosam430]

PB²=1²+2²=5--->PB=√5 ---> Potencia de P respecto a la circunferencia =1*1+2=3 =PQ*PB=PQ*√5---> PQ=3√5/5 ---> QB²=(√5)²-(3√5/5)²---> QB=4√5/5 ---> Área PQB=PQ*QB/2=6/5. Gracias y saludos

[@michaeldoerr5810]

The area is 6/5 units square. I really appreciate this review of how a circle theorem leads up to similarity by HL in the first method. I think that I finally got the triple understanding of how similarity by HL usually works: First you take the 90 degree angle than the common angle, the triangles that are subtended by the 90 degree angle are similar and at the 4:00 mark, it is clear that the first two letters of the 90 degree angle pair and the first and last letters of the common angle pair are set equal. Then the proportions are substituted to get x^2 and then you subtract from the larger right triangle from the isosceles triangle. As for the second method it makes use of a circle theorem and chord theorem and then Pythagorean Theorem. Both methods make use of the Pythagorean Theorem. Please check out this comment.

[@RealQinnMalloryu4]

(2^2=4 180°ABC/4 =40.20 2^20.2^10 2^2^10.2^2^5 1^12^5.1^1^1 1^12^1 2^1 (ABC ➖ 2ABC+1) .

[@murdock5537]

∆MQB → BM = 4 = MO + BO = 2 + 2; PO = 1; sin⁡(MOP) = 1 → PM = PB = √5; BQ = k; PQ = m QMB = PMB = MBP = δ → sin⁡(δ) = √5/5 → cos⁡(δ) = 2√5/5 = (√5 + m)/4 → m = 3√5/5 → MQ = 8√5/5 → area ∆ BQM = (1/2)sin⁡(δ)4(√5 + m) = 16/5 → area ∆ BQP = (1/5)(16 - 10) = 6/5

[@Emerson_Brasil]

*_3° Método:_* Sejam PQ=a e QB=b. Note que a soma dos ângulos opostos do quadrilátero QBOP é 180°, portanto, QBOP é inscritível. Segue do teorema de Pitolomeu: *PQ×OB + QB×OP=PB×OQ. (1)* Por Pitágoras, no ∆PBO: PB²=OB²+PO²=4+1→PB=√5. observe que agora, que OQ=raio do círculo=2. Usando a equação (1), obtemos: 2a+b=2√5→ *b=2√5-2a* Usando Pitágoras no ∆PQB: PB²=PQ²+QB². Assim, (√5)²=a²+b² 5=a²+(2√5 - 2a)² 5=a²+20 - 8√5a + 4a² 5a² - 8√5a + 15=0. Usando a fórmula para uma equação de 2° grau, temos: ∆=320 - 300=20→√∆=2√5. Daí, a=(8√5 ± 2√5)/10. Assim, a=√5 ou a=3√5/5. Se a=√5, então b=0, absurdo! Pois a,b >0, já que são lados de um triângulo. Logo, a=3√5/5, consequentemente, b=2√5 - 6√5/5=4√5/5. A área do triângulo desejado é dado por a×b/2, isto é, (3√5/5 × 4√5/5)/2=(12×5/25)/2. Finalmente, *6/5.*

[@Emerson_Brasil]

*_4° Método:_* Sejam PQ=a e QB=b. Note que a soma dos ângulos opostos do quadrilátero QBOP é 180°, portanto, QBOP é inscritível. Traçando a diagonal OQ, temos que os ângulos ∠BOQ=∠BPQ=α (ARCOS CAPAZ). Usando lei do cosseno no ∆BOQ: QB²=QO²+OB² - 2QO×OBcos α b²=2²+2²-2×2×2cos α *b²= 8 - 8cos α (1)* Por Pitágoras, no ∆PBO: PB²=OB²+PO²=4+1→PB=√5. Além disso, cos α= PQ/PB→ *cos α=a/√5 (2)* Usando Pitágoras no ∆PQB: PB²=PQ²+QB². Assim, (√5)²=a²+b²→ *b²= 5 - a² (3)* Substituindo (2) e (3) em (1): 5 - a²= 8 - 8a/√5 a² - 8√5a/5+3=0 × (5) 5a² - 8√5a + 15=0. Usando a fórmula para uma equação de 2° grau, temos: ∆=320 - 300=20→√∆=2√5. Daí, a=(8√5 ± 2√5)/10. Assim, a=√5 ou a=3√5/5. Caso seja a=√5, por (3), teremos b=0, absurdo! Pois a,b >0, já que são lados de um triângulo. Só nos restou a=3√5/5, consequentemente, por (3), temos: b²=5 - (3√5/5)²=5 - 9/5=16/5 b= √(16/5) →b=4/√5. A área do triângulo desejado é dado por (a×b)/2, isto é, (3√5/5 × 4/√5)/2=(12/5)/2. Finalmente, *6/5.*

[@giuseppemalaguti435]

QPB=α..legge del coseno 4=1+(√5cosα)^2-2*1*√5cosαcos(α+arctg2)...sin2α=24/25..Α=(1/2)√5cosα√5sinα=(1/4)5sin2α=6*5/25=6/5

[@yakupbuyankara5903]

6/5=1.2

[@Nobody-tu5wt]

6/5

[@himadrikhanra7463]

1 square unit ?

[@jimlocke9320]

At about 4:10, Math Booster has found that ΔMOP and ΔMQB are similar. The hypotenuse MB of ΔMQB has length 4. From the Pythagorean theorem, the hypotenuse of ΔMOP has length √5. The areas of similar triangles are related by the square of the ratio of corresponding sides. In this case, take the ratio of the hypotenuses, which is 4/(√5) and square it, giving 16/5. The area of ΔMOP = (1/2)(bh) = (1/2)(2)(1) = 1. Therefore, the area of ΔMQB is (16/5)(1) = 16/5. ΔMOP and ΔBOP are congruent by side - angle - side, so ΔBOP also has area 1. Area ΔBQP = ΔMQB - ΔMOP - ΔBOP = 16/5 - 1 - 1 = 6/5, as Math Booster also found.

[@jairoeveliogordillomarin5780]

Aplicando teorema de Pitágoras: PB= √5. Tangente OBP= 1/2. Entonces: ángulo OBP= 26.57°. Ángulo PBQ: 180=26.57+90+26.57+PBQ Entonces: ángulo PBQ= 36.86°. Seno 36.86=PQ/√5. PQ= 1.34 u. Coseno 36.86=PQ/√5= 1.79 u Área triángulo PQB= BQ×QP= (1.34×1.79)/2= 1.1993 u².

[@batmunkhenkhbaatar9061]

I could solve it by myself

[@Emerson_Brasil]

@batmunkhenkbaatar9061 That's exactly the purpose of the videos! If you can solve it yourself, my dear friend!