Find the blue shaded area of the triangle inside a quarter circle | Maths Olympiad Geometry Problem

Find the blue shaded area of the triangle inside a quarter circle | Maths Olympiad Geometry Problem
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[@KaushikLaskarT.I.M.E.Mumbai]

Since PQBC is a cyclic quadrilateral, using Ptolemy's theorem we get X.R + X.(R-2) = Sqrt(2).X.R Hence R = 2+ Sqrt(2) Area of QBC = 1/2.R.(R-2) = 1+Sqrt(2)

[@julioguerra5092]

How do you prove it is a cyclic quadrilateral?

[@احمدمحمد-ت8ط9ج]

Snglep+angleb=90+90=180 DEGREES

[@hongningsuen1348]

Definitely this is the best method as it needs minimal construction (diagonal BD only) and algebra (X in all 3 terms of equation can be cancelled out).

[@Okkk517]

By observing that 🔼BDQ ~ 🔼CDP we obtain the following equation: (R-2)/R=x/(x+sqrt(2)*x)=1/(1+sqr(2)). Solving for R we get R=2+sqrt(2). Therefore, the required area is equal to R*(R-2)/2=1+sqrt(2).

[@stgshaurya8573]

can i ask you something that Why u have Written x+root2x . i think it should be root2 only . explain pls if u have time Reply

[@Okkk517]

Triangles BDQ and CDP are similar. Then, QB/BD=CP/DP . Moreover, CQ=DQ=sqrt(2)*x (symmetry). Therefore, (R-2)/R=CP/(PQ+DQ)=x/(x+sqrt(2)*x).

[@StaR-uw3dc]

Nice and refined solution. Other approach: First equation: x^2 = R^2+2-2R Other variant of the second equation: In triangle CPD we have: CP^2+PD^2=CD^2 => x^2 + (x+sqrt(2)x)^2 = (2R)^2 x^2(1+1+2sqrt(2)+2) = 4R^2 x^2(4+2sqrt(2)) = 4R^2 x^2 = 4R^2/(4+2sqrt(2)) = 2R^2/(2+sqrt(2)) = 2R^2(2-sqrt(2))/[(2+sqrt(2))(2-sqrt(2))] = 2R^2(2-sqrt(2))/(4-2) = R^2(2-sqrt(2) Finally x^2 = R^2(2-sqrt(2)) Comparing first and second equation we get quadratic in R: R^2+2-2R = R^2(2-sqrt(2)) R^2(1-sqrt(2)) +2R -2 = 0 with roots R1=sqrt(2) and R2=2+sqrt(2). Due to R-2>0 the only solution for R is R2=2+sqrt(2).

[@quigonkenny]

Mirror arc CA about AB to create 90° arc AT. Extend PQ and CB to intersect at T. By Thales' Theorem, as ∠P is a 90° angle on a circumference, C and T are ends of the diameter of the circle. By observation, as CT is the diameter, BC and BT are radius r and QC = QT. Thus ∆QBC ≅ ∆TBQ. Let s be the side length of the two matching sides of the isosceles right triangle. Triangle ∆CPQ: CP² + PQ² = QC² s² + s² = QC² QC² = 2s² QC = √(2s²) = √2s By observation, as ∠T is shared, right angle triangles ∆CPT and ∆TBQ (as well as ∆QBC) are similar. Triangle ∆TBQ: QB/BT = CP/PT (r-2)/r = s/(s+√2s) = 1/(1+√2) r = (r-2)(1+√2) r = r + √2r - 2 - 2√2 √2r = 2 + 2√2 r = (2√2+2)/√2 r = 2 + √2 Triangle ∆QBC: A = bh/2 = r(r-2)/2 A = (2+√2)(2+√2-2)/2 A = (2+√2)(√2)/2 A = (2√2+2)/2 = 1 + √2

[@vacuumcarexpo]

This is tougher than I expected.

[@xz1891]

Bcz the method is too clumsy

[@jasondai857]

This will a simple way: ∵ ∠CBQ=CPQ=90° ∴CPQB on a circle and CQ is the diameter. ∵ CPQ=90° and CP=PQ ∴ ∠CQP=45° tie PB, ∠CBP=∠CQP=45° ∠ABP=∠CBP=45° ∴P is tthe midle point of arc AC, ∵BP=BC=R ∴ △BPC is aisosceles triangle ∠PCB=∠CPB=(180-45)/2=67.5° ∠BCQ=22.5° ∠CQB=67.5° ∠AQP=180-67.5-45=67.5° Let PE⊥AB，E on AB. tie AP, ∵ P is tthe midle point of arc AC ∴AP=PC=PQ △APQ is aisosceles triangle ∴E will be midle point of AQ ∵ AQ=2 ∴ EQ=1 ∵ PE⊥AB, ∠ABP=45° ∴ ∠BPE=45° ∴ BP/PE=√2 ∵∠AQP=67.5° ∴ ∠EPQ=90-67.5=22.5° PQ is angle bisector of ∠BPE ∴ BP/PE=BQ/EQ=√2 BQ=√2EQ=√2 ∴ AB=BC=2+√2 Blue Area = BCxBQ/2=(2+√2)x√2/2=√2+1

[@honestadministrator]

Extension of PO meets D at circumference. Angle PCD is right angle so CD is diameter. ∆ COB is congruent to ∆ DOB and both of them are similar to ∆ DCP CB / DP = CO / DC 2 R^2 = DP x CO = (DO + OC) CO

[@じーちゃんねる-v4n]

If the radius is r　∠BCP=∠BPC=θ　BC=BP=r PC=PQ=2rcosθ BQ=r-2 PC=2rcosθ QC=2√2rcosθ 　 From Tremy's theorem 2r^2cosθ+2r(r-2) cosθ=2√2r^2 cosθ　 ∴r+(r-2)=√2r ∴r=2+√2　∴S=(1/2)√2(2+√2)=1+√2

[@biaohan4358]

Since you drew the lines to D there's a very easy way to solve the problem without the need to use x and the other semicircle. Basically angle CQP is pi/4 obviously, so angle CQD is 3*pi/4, and angle CQB is half of that (3*pi/8 or 62.5 degree). We only need to know the tangent of this angle. From the half angle equation tan(3pi/8)=+/- sqrt((1-cos(3pi/4))/(1+cos(3pi/4)))=sqrt(2)+1 (obviously it's positive). Therefore R/(R-2)=sqrt(2)+1 it would be just a few seconds to get R=2+sqrt(2).

[@saidalazn]

Considering that PBC=45, PBQ= 45 we find: In △PBC: X^2 = R^2 + R^2 - 2.R.R.cos45 In △PBQ X^2 = R^2 +(R-2)^2 -2.R.(R-2).cos45 Thus: R^2 + R^2 - 2.R.R.cos45 = R^2 +(R-2)^2 -2.R.(R-2).cos45 Solving the equation for R we can find the required area.

[@murdock5537]

Great! Many thanks, Sir! tan⁡(φ) = x/(x + x√2) = (r - 2)/r = √2 - 1 → r(√2 - 1) = r - 2 → r = 2 + √2 → blue area = 1 + √2 btw: x = (√2)(√(2 + √2)) → r = 2 + √2 → tan⁡(φ) = (r - 2)/r = (1/2)(√2)(2 - √2) = √2 - 1 → tan^2(φ) = 3 - 2√2 →1 - tan^2(φ) = 2(√2 - 1) → tan⁡(2φ) = (2tan⁡(φ))/(1 - tan^2(φ)) = 1 → 2φ = 45° = PBC

[@zakizaki1197]

Great job

[@PaulRM329]

Could you explain the introduction of coefficients "a" and "b". I think, if I've read your working correctly, then "a" and "b" correspond to the of coefficients of "R" which are "2" and "2.2^(1/2). However, when you calculate the product of "a" and "b", the coefficient of "R^2" is introduced. It's left me somewhat confused. Cheers.

[@MathBooster]

For any quadratic equation ax²+bx+c=0 , to make factors, we need to write b as the sum of two numbers suppose p and q (b=p+q) and p•q it should be equal to a•c For example : 2x²+7x+6=0 , it is 2x²+4x+3x+6=0 (now you can factor it) 3x²-8x+5=0 it is 3x²-3x-5x+5=0 (now you can factor it)

[@jimlocke9320]

@@MathBooster I think the quadratic equation solution may be more straightforward than factoring. For aR²+bR+c=0, a = 1, b = -(2+2√2), and c = 2+2√2. b²=(2+2√2)²=(4+8√2+8)=12+8√2. 4ac=8+8√2, so b²-4ac = 12+8√2-(8+8√2)=4 and √(b²-4ac) = √4 = 2. So the 2 solutions from R=(-b+√(b²-4ac))/2a and R=(-b-√(b²-4ac))/2a are R=((2+2√2)+2)/2 = 2+√2 and R=((2+2√2)-2)/2 = √2, as you found by factoring.

[@claudeabraham2347]

I love it!

[@MarieAnne.]

Here is another method I used. Let BC = AB = R, then QB = AB − AQ = 2 − R Extend CB and PQ to point D (as done in video) to get right △CPD Since ∠P = 90°, then this angle must be subtended by a diameter of the circle. Therefore, CD is a diameter of circle, and we draw semi-circle through C, P, A, and D. By symmetry, △CBQ ≅ △DBQ, so ∠BQC = ∠BQD = θ Triangle CPQ is isosceles right triangle with CP = PQ. Therefore, ∠PCQ = ∠PQC = 45° Since PQD is a straight line, then ∠PQD = 180° ∠PQC + ∠CQB + ∠BQD = 180° 45° + θ + θ = 180° 2θ = 135° Using trig identity, we get: tan θ = (1 − cos 2θ) / (sin 2θ) tan θ = (1 − cos 135°) / (sin 135°) tan θ = (1 + 1/√2) / (1/√2) tan θ = √2 + 1 In △CBQ we get: tan(∠BQC) = BC/BQ tan θ = R/(R−2) √2 + 1 = R / (R−2) √2 R + R − 2√2 − 2 = R √2 R = 2√2 + 2 R = 2 + √2 Now we can calculate area of blue triangle: BC = R = 2 + √2 BQ = R − 2 = √2 Area = 1/2 × √2 × (2 + √2) = 1/√2 × (2 + √2) = *√2 + 1*

[@EnnioPiovesan]

Consider the figure drawn at minute 3:53. Let's call QCB=β. In the right triangle PCD we have PCD=45°+β (1). Triangles QCB and QBD are equivalent so QDB=QCB=β. (2). Using (1) and (2) we have 45°+β +β = 90° --> β=45°/2. But PDC is the circumferential angle subtended by the chord PC, hence the angle at the center of the circumference PBC=45°. As a consequence PBA=45° so PC=PA. But PC=PQ so also PA=PQ, hence triangle APQ is isosceles with the base AQ. Let's draw the perpendicular to AQ through P in H. We have AH=HQ=1. Now consider the right triangles PHQ and CBQ; they are similar because both right triangles and CQB=BQD (as corresponding angles in the equivalent triangles QCB and QBD) = PQH (opposite angle to BQD). So we can write: QC:PQ=QB:HQ --> √2=(R-2)/1 --> R=2+√2. [...]

[@TrebolManger]

Summary Words HypTEq = HypTBlue ∧ ARect - ASq = ARmb Equations 2 2 2 2 (2 r) (2 (r - 2)) (√2 x) = r + (r - 2) ∧ x (x + √2 x) - x = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 2 Algebra System 2X2 x = √(2 √2 + 4) ∧ r = √2 + 2 r (r - 2) ATBlue = ⎯⎯⎯⎯⎯ 2 Substitution (√2 + 2) ((√2 + 2) - 2) ATBlue = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 2 ATBlue = √2 + 1

[@xz1891]

Your method is too algebraic and, tbh, redundant for such a geometric problem. See China grade 6 way of mine... Link PA, PB, note PB=BC, let be r. Now look first at quad PQBC, the 4 pts are co-circular, thus ∠QBP =∠PCQ=45°, thus pt P is at halfway of AC arc, now Draw PM, PN, respectively perpendicular to AB, and BC. Thus, CP=AP, PM=PN, so, ∆PAM≡∆PQM≡∆PCN (congruence), thus AM=QM=CN=1, then we have LINEAR equation (No quadratic whatsoever) r-1 = r/√2, solve, r = 2+√2 So shaded area = (2+√2)*√2 /2= √2 +1 #😊

[@arongamingff5m566]

Easy one❤️❤️❤️

[@luiscalpo3813]

1/2 is one over two and not one by two.

[@JeanBaptisteEmanuelZorg]

Can't I just use a ruler and compute the field of the triangle? Or is it yet another mathematics imaginary problem that prohibts you to use any tool(s)?

[@rohitmadashri7250]

Of course you can. I sprayed rat poison on the blue area and threw rats, calculated the number of dead ones to get the area. Another way would be to use quantum computing to solve this area. If nothing works, I just pick up Mills and Boon and bide my time.

[@spacer999]

You can find R much faster using similar triangle: QBC ~ CPD. (R-2)/R = x/(x + √2x) = 1/(1+√2) --> Solve for R gives R = √2 + 2

[@stgshaurya8573]

can i ask you something that Why u have Written x+root2x . i think it should be root2 only . explain pls if u have time

[@thuyngoc2542]

Chưa g/t vì sao AQ.QM=PQ.QD

[@mariamalfaky128]

I want to meet you llove you

[@rajendranarayanshukla2777]

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