A Very Nice Math Olympiad Geometry Problem | Given a circle inside a quarter circle

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A Very Nice Math Olympiad Geometry Problem | Given a circle inside a quarter circle
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Пікірлер: 37

@MarieAnne. 11 ай бұрын

Once we've constructed △CDE (as shown at 5:56) we can find CE using Pythagorean Theorem: CE = √(DE² − CD²) = √(6² − 2²) = √32 = 4√2 sin 2θ = CE/DE = (4√2)/6 = 2√2/3 cos 2θ = CD/DE = 2/6 = 1/3 tan 2θ = CE/CD = (4√2)/2 = 2√2 We can get tan θ using half-angle identity: tan θ = (1 − cos 2θ) / sin 2θ = (1 − 1/3) / (2√2/3) = (3−1)/(2√2) = 1/√2 In △OBE tan2θ = OE/OB → OB = OE/tan2θ = 3/(2√2) In △ABP tan θ = AP/AB → AB = AP/tan θ = r/(1/√2) = r√2 In △OAP AP = r OA = OB + AB = 3/(2√2) + r√2 = (3+4r)/(2√2) OP = 3−r Using Pythagorean theorem, we get: r² + ((3+4r)/(2√2))² = (3−r)² r² + (16r²+24r+9)/8 = 9 − 6r+ r² (16r²+24r+9)/8 = 9 − 6r 16r² + 24r + 9 = 72 − 48r 16r² + 72r − 63 = 0 16r² − 12r + 84r − 63 = 0 4r (4r − 3) + 21 (4r − 3) = 0 (4r + 21) (4r − 3) = 0 Since r > 0, then *r = 3/4*

@اممدنحمظ Жыл бұрын

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

@zsboya Жыл бұрын

@marioalb9726 Жыл бұрын

Labelling OB=a, BD=b sin α = 2/6 α = 19,4712° b = 3/ cosα = b = 3,18198 cm Similarity of triangles: a/b = 2/6 a = 1,06066 cm α + 2θ = 90° θ = 35,2644° Tangent secant theorem: (a + r/tanθ)² = R. (R-2r) (a + r/tanθ)² = 3(3-2r) (a + r/tanθ)² = 9-6r Clearing 'r' : r = 0.75 cm ( Solved √ )

@marioalb9726 Жыл бұрын

sin α = ½ 2 / 3 = 1/3 This right triangle will be named ODC' C' is midpoint of CD, and C'D=1 Rotate triangle ODC' about O, until align OD and OP: sin α = 1/3 = r / (R-r) 3.r = R-r 4r = R r = R/4 = 3/4 = 0.75 cm (.Solved √ )

@jarikosonen4079 Жыл бұрын

ODC' ~ EDC => DC'=DC/(6/3)=1... Then how the rest? Not sure if this is proven.

@marioalb9726 Жыл бұрын

@@jarikosonen4079 Very challenging this exercise of Math Booster. Nowadays, the prove is drawing to scale, in computer, in CAD. Enough provement !! I'm engineer, not mathematician

@jarikosonen4079 Жыл бұрын

@marioalb9726 It looks good to check it using actual drawing.

@zsboya 11 ай бұрын

We need to prove that the right triangles CDE and APO are similar. In this case, we can calculate "r" very easily with ratio pairs. But how? \ a =

@tunneloflight 11 ай бұрын

After a little head scratching …. Elegantly done.

@marioalb9726 Жыл бұрын

Being 'α' the angle between OD and CB, and also between OD and BD , this means α = Angle BDO sin α = 2 / 6 = 1/3 Also α is angle POA, due to triangle BDO is similar to triangle POA r / (R-r) = 1/3 3r = R-r 4r = R r = ¼R = 3/4 = 0.75 cm ( Solved √ )

@erskinelowe8360 Жыл бұрын

where is the proof ta triangle BDO IS SIMILAR TO TRIANGLE POA?

@zsboya Жыл бұрын

Hi! BDO = BEO = α is understandable since triangle BDE is isosceles. But how does it follow that the triangles BDO and POA are similar?

@marioalb9726 Жыл бұрын

@@zsboya As you can see, the result is correct, r = 0.75 cm , same result as video and other comments. Those triangles are similar !! However, It seems that this exercise was created somewhat misleading. Those triangles are similar only with these dimensions, both angles have same value, but they DON'T HAVE TO BE. So, you are right !!! I was wrong !!! I fell into the trap !! See below other solution:

@marioalb9726 Жыл бұрын

Labelling OB=a, BD=b sin α = 2/6 α = 19,4712° b = 3/ cos α b = 3,18198 cm Similarity of triangles: a/b = 2/6 a = 1,06066 cm α + 2θ = 90° θ = 35,2644° Tangent secant theorem: (a + r/tanθ)² = R. (R-2r) (a + r/tanθ)² = 3(3-2r) (a + r/tanθ)² = 9-6r Clearing 'r' : r = 0.75 cm ( Solved √ )

@zsboya Жыл бұрын

I think it is likely that this would be a simple proof of the task, only that it would be necessary to somehow prove that the triangles CDE and APO are similar.

@quigonkenny 4 ай бұрын

Extend DO 3 units to E. Extend the arc of the quarter circle around to E to create a semicircle. Extend CB to E. as D and E are ends of a diameter and ∠C = 90, CB is colinear with CE. Triangle ∆DCE: DC² + CE² = ED² 2² + CE² = 6² CE² = 36 - 4 = 32 CE = √32 = 4√2 As ∠EOB = ∠DCE = 90° and ∠BEO = ∠CED, ∆EOB and ∆DCE are similar. OB/OE = DC/CE OB/3 = 2/4√2 = 1/2√2 OB = 3/2√2 Let P be the center of the circle, T be the point of tangency between circle P and the circumference of quarter circle O, and V be the point of tangency between circle P and CB. Draw PA, PV, and PB. AB and VB are tangents from circle P intersecting at B, so AB = VB. As PA = PV = r, and PB is shared, ∆PAB and ∆BVP are congruent. Let ∠ABP = ∠PBV = α. As ∠ABV thus equals 2α, and OA and VE are straight lines, ∠OBE = 2α as well, being a vertical angle of ∠ABV. As ∆EOB and ∆DCE are similar, ∠EDC = ∠OBE = 2α. tan(2α) = CE/DC = 4√2/2 = 2√2 sin(2α) = CE/ED = 4√2/6 = 2√2/3 cos(2α) = DC/ED = 2/6 = 1/3 tan(2α/2) = sin(2α)/(1+cos(2α)) tan(α) = (2√2/3)/(1+1/3) tan(α) = (2√2/3)/(4/3) tan(α) = 2√2/4 = 1/√2 Triangle ∆PAB: tan(α) = PA/AB 1/√2 = r/AB AB = √2r OA = AB + OB OA = √2r + 3/2√2 OA = (4r+3)/2√2 Draw radius OT. As the centers of two circles are colinear with their point of tangency, OT passes through P. As PT = r and OT = 3, OP = 3-r. Triangle ∆PAO: PA² + OA² = OP² r² + ((4r+3)/2√2)² = (3-r)² r² + (16r²+24r+9)/8 = 9 - 6r + r² 16r² + 24r + 9 = 72 - 48r 16r² + 72r - 63 = 0 16r² + 84r - 12r - 63 = 0 4r(4r+21) - 3(4r+21) = 0 (4r+21)(4r-3) = 0 r = -21/4 ❌ (r ≥ 0) | r = 3/4 ✓

@parvinkumarmalhotra3644 4 ай бұрын

😅😅😮

@tunneloflight 11 ай бұрын

Btw - your solution is excellent. You may be able to shorten it a bit by bisecting OC (C’ as others suggest), then finding the angle CDO (2 theta) by solving the triangle C’DO, as OC’D must be a right angle. Cos(2 theta) = 1/3 or OC’ = root(9-1) = 2*root(2), so tan(2 theta) = 2*root(2).

@EnnioPiovesan Жыл бұрын

As usual I want to try to avoid trigonometry. We assume and accept the fact that: BO=(3/4)√2, CE=4√2, CDE=2ϑ, ABP=ϑ. Now draw from O the perpendicular to the chord CE, which intersects CE itself at R and the (big) circumference at H. We observe that ORE is similar to DCE with a similarity ratio of 1:2, so that RO=(1/2)CD=1. Consequently HR=HO-RO = 3-1=2. Since the perpendicular to the chord through the center divides the chord itself into two equal parts, CR=RE and therefore RE=(1/2)CE)=2√2. Applying the Pythagorean theorem to the triangle HRE we have: HE=√(HR²+RE²)=√(4+8)=2√3.Applying the Pythagorean theorem again to the triangle DHE we have: HD=√(DE²-HE²)= √(36-12)=2√2√3. The ratio HE:HD is therefore 2√3:2√3√2= √2/2. We observe that the two arcs HE and HC are equal and therefore subtend equal angles to the circumference, so (in view of the fact that the total arc CE subtends CDE=2ϑ) HDE=HCD=ϑ. Consequently the right triangles DHE and PAB are similar and therefore HE:HD=PA:AB=r/AB=√2/2 for which AB=r√2. Tracing the ray from O passing through P until it meets the greater circumference in M and the smaller in G, we can write OA²=OM OG --> (AB+BO)²=OM OG --> (r√2+ (3/4)√2)² = 3(3-2r) --> ... 16r²+72r-63=0, which finally yields r=3/4.

@zsboya Жыл бұрын

Hi! Nice derivation, but it's very long. By the way, you wrote it wrong in one place: HDE = HDC (Not HCD) = ϑ.

@EnnioPiovesan Жыл бұрын

You are right. Thank you.@@zsboya

@ציוןעמרני-ס1פ 7 ай бұрын

In the solution, you assumed that O,P,M reside on the SAME line OM. That enables you to conclude that OP=3-r How can you prove that?

@User-jr7vf 5 ай бұрын

Not gonna lie, this one I had to solve in a calculator. And it took me about 2 hours (!!!) to properly set up the problem. As usual I only check the solution in the video after I found the answer myself, and it turns out the method by Math booster is similar to what I used, but her method is easier computation-wise.

@ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μ9μ Жыл бұрын

Στο σημείο που θέλετε να δείξετε ότι

@spacer999 Жыл бұрын

If you rotate triangle OPA about O so that OP aligns to OD, it's obvious that triangles OPA is similar to triangle FDC. So OP/PA = FD/CD --> (3-r)/r = 6/2 --> 3-r = 3r --> 3 = 4r --> r = 3/4.

@MathBooster Жыл бұрын

Nice But you need to prove that the two triangles are similar. One angle is 90° that is equal in both triangles so you need to show that any second angle is also equal in those two triangles.

@spacer999 Жыл бұрын

@MathBooster you are right. I wonder if there is a simple way to show say angle AOP = angle DFC.