merci sire , excellent exercice

@PreMath

29 күн бұрын

Excellent! You are very welcome! Thanks for the feedback ❤️🙏

tan α = Rcos45°/R = 1/√2 α = 35,26° 2R = 32 / cos α R = 19,596 cm A = ½R² A = 192 cm² ( Solved √ )

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

In the third step, instead of considering the similar triangles, extend OP to become a diameter of the circle and then use the intersecting chords theorem: (32 - a√3)a√3 = (a√2 + a)(a√2 - a) = a² => a = 8√3

@PreMath

29 күн бұрын

Good job! Thanks for the feedback ❤️

Nice Solution for the rigorous problem. Thank you Sir.

Brilliant use of Similar triangles 📐 in the end. Can we know the name of the app/software with which you are writing ✍️ electronically on the computer screen?

Let the radius of semicircle O be r and the side length of square OPDE be s. Draw radius OD. As OD is also the diagonal of square OPDE, this gives us a direct relationship between s and r via Pythagoras: Triangle ∆OPD: OP² + PD² = OD² s² + s² = r² 2s² = r² s² = r²/2 s = √(r²/2) = r/√2 --- [1] r = √2s --- [2] Draw CB. By Thales' Theorem, as A and B are opposite ends of a diameter and C is a point on the circumference, then ∠BCA = 90°. As ∠AOP = 90° as well, and ∠PAO is common, triangles ∆AOP and ∆BCA are similar. Triangle ∆BCA: BC/CA = OP/OA BC/32 = s/r = (r/√2)/r --- [1] BC = 32/√2 = 16√2 BC² + CA² = AB² (16√2)² + 32² = (2r)² = (2(√2s))² --- [2] 8s² = 512 + 1024 = 1536 s² = 1536/8 = 192 sq units

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Recognizing that AO is the same as the diagonal of the square, that is, both are radii, I took that arctan of PAO as a/(a*2^.5) which was 1/2^5. That angle was 35.26438... So the cosine of this angle is 32 divided by 2*a*2^.5 (which is the diameter). Solve for a which is 13.85640646... Then square that number to equal 192.

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Be R the radius of the circle and t = angleBAC In triangle AOB: OP = R/sqrt(2) and AO = R Then AP^2 = OP^2 +AO^2 = (R^2)/2 + R^2 = (3/2).R^2 and AP = sqrt(3/2).R Then cos(t) = AO/AP = R/(sqrt(3/2).R) =sqrt(2/3) In triangle ABC: cos(t) = AC/AB, so: sqrt(2/3) = 32/(2.R) and R = 16.sqrt(3/2) The area of the square is (R/sqrt(2))^2 =(R^2)/2 = (256.(3/2))/2 = 64.3 = 192

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Diagonal of the square = radius of circle =a√2 AP =√[ (a√2)^2 +a^2]=a√3 DP is extended to T to get a chord. Now intersecting chord theorem states that DP *PT =AP *PC -- (1) (DP =PT =a) AP=a√3 PC= 32 - a√3 Now from equation No. 1 we get a^2= a √3(32 - a√3) > 4a^2=32a√3 > 4a =32√3 > a=8√3 a^2=(8√3)^2=64*3=192 Area of square = 192 sq units Comment please

@PreMath

29 күн бұрын

Good job! Thanks for sharing ❤️

with the help of you, i can❤

@PreMath

29 күн бұрын

Super! Thanks for the feedback ❤️

R=x\/2, ВС/32=х/х\/2, ВС=32/\/2=16\/2, (2R)^2=32^2+(16\/2)^2, (2x\/2)^2=1024+512, 8x^2=1536, x^2=1536/8, x^2=192.

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

I've learned 3 methods from the video and comments: After r (radius) = sqrt2 x a (side of square) by Pythagoras theorem 1. Using Thales theorem and then similar triangles to get a 2. Using sides of right-angled triangle to get cosBAC then Thales theorem to get r and then a 3. Using intersecting chords theorem to get a

@PreMath

29 күн бұрын

Super! Glad to hear that! Thanks for the feedback ❤️

Okay, I'm dazzled.

One more solution: 1. Let a= OP=PD; x=AP; PC = 32-x; 2. Let us extend DP to the left to intersect with the circle at point N => PN = PD = a; 3. PN*PD = x*(32-x) => a^2 = x*(32-x); (1) 4. x^2 = r^2 + a^2; (2) 5. r^2 = 2*a^2; (3) 6. system of equations {(1),(2),(3)} 7. let us put (3) into (2): x^2 = 3*a^2 => x = a*sqrt(3); (4) 8. Let us put (4) into (1): a^2=a*sqrt(3)*(32-a*sqrt(3)); as a!=0 lets devide both sides by a a = 32*sqrt(3) - 3*a; 4a = 32*sqrt(3); a= 8*sqrt(3); Ablue = a^2 = (8*sqrt(3))^2 = 64*3 = 192 sq units.

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Good morning sir I like it

@PreMath

29 күн бұрын

Hello dear❤️ Thanks for liking ❤️

Cos(A)=√(2/3). Опускаем перпендикуляр из центра окружности на AC. Делит хорду пополам. а√2=16/(√2/3). а=8√3.

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

PreMath likes square roots a lot. And when it should and when it shouldn't. Let's ban square roots! OD^2=2a^2. AP^2=3a^2. (AB^2)/(AC^2)=(AP^2)/(AO)^2. (8a^2)/32^2=(3a^2)/2a^2. a^2/64=2. a^2=192 square units.😎

@PreMath

29 күн бұрын

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3rd alternative solution: 1/ Just assume that a=1 so r= sqrt2 and AP= sqrt3 and area of blue square= 1 2/ The quadrilateral PCBO is cyclic so, APxAC=AOxAB-->ACxsqrt3=sqrt2x2sqrt2=4--> AC=4/sqrt3 Now if AC = 32, the area of the blue square = 1x sq((32/(4/sqrt3))=64x3= 192 sq units😊

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Let's find the area: . .. ... .... ..... Since OEDP is a square, all interior angles are right angles. Therefore the triangle ODE is a right triangle and we can apply the Pythagorean theorem. With r being the radius of the semicircle and s being the side length of the square we obtain: OD² = OE² + DE² r² = s² + s² = 2*s² According to Thales theorem the triangle ABC is a right triangle. The triangle OAP is also a right triangle. Now let's have a look at the interior angles of these two triangles: ∠AOP = ∠ACB = 90° ∠OAP = ∠BAC = α ∠APO = ∠ABC = β So these two triangles are similar and we can conclude: AO/AP = AC/AB r/AP = 32/(2*r) r/AP = 16/r AP/r = r/16 ⇒ AP = r²/16 Now we can apply the Pythagorean theorem to the right triangle OAP in order to obtain the area of the square: AP² = AO² + OP² (r²/16)² = r² + s² (2*s²/16)² = 2*s² + s² (s²/8)² = 3*s² s⁴/64 = 3*s² Since s=0 is not a useful solution, the only useful solution is: A(square) = s² = 3*64 = 192 Best regards from Germany

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Very nice! ∆ ABC → AB = 2r = AO + BO; AC = 32; ∎EDPO → PO = EO = DE = DP = a → DO = AO = BO = a√2 = r; sin⁡(AOP) = 1 → PAO = δ → PO = a; AO = a√2 → AP = a√3 → cos⁡(δ) = a√2/a√3 = √6/3 = 32/2a√2 → a = 8√3 → ∎EDPO = 64(3) = 192

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️ Thanks for the feedback ❤️

OBDP is a square So OB=BD=DP=PO=x And OD is Radius of semicircle =x√2 ∆ AOP~ ∆ ACB x/x√2=BC/32 1/√2=BC/32 So BC=16√2 In ∆ ABC (16√2)^2+(32)^2=(2x√2)^2 So x=8√3 Blue square area =(8√3)^2=192 square units.❤❤❤

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Thank you so much! My first attempt was almost the same as yours😅! Here is my 2nd approach: Let a and r be the side of the square and the radius of the circle respectively. We have: r= a. sqrt2 So tan angle CAB= OP/OA=a/r=1/(sqrt2) Note that the angle ACB= 90 degrees so, BC/AC=tan(CAB)=1/sqrt2 --> BC=16sqrt2 By using the Pythagorean theorem sqAC+sqBC=sqAB -> sq32+sq(16sqrt2)=sq(2r)=sq(2.a.sqrt2) --> 8 sqa=512+1024 Area= sq a=192 sq units

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

32/2r=r/√(r^2+l^2)...l,lato del quadrato..con r=√2l..sostiisco, risulta l=8√3...A=192

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Ok, muy bien

@PreMath

29 күн бұрын

Excellent! Thanks for the feedback ❤️

PO=OE=ED=DP=x OD=AO=OB=√2x ⊿AOP∞⊿ACB BC=32/√2=16√2 AB=2√2x (16√2)²+32²=(2√2x)² 512+1024=8x² 8x²=1536 x² = Blue Square area = 192

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

chord theorem wouldmake it simpler

Congratulations for the question sir!!

@PreMath

29 күн бұрын

Excellent! Glad to hear that! Thanks for the feedback ❤️

I guess when I think outside the box I'm really looking for a distant function I project onto the space in question. 🤔

@PreMath

29 күн бұрын

Excellent! 😀 Thanks for the feedback ❤️

tan a=1/sqrt(2), so cos a=1/sqrt(3/2), chord length=2 r cos a=2sqrt(2/3)r=32, r=16sqrt(3/2), the area is r^2/2=3×128=384.😊

@PreMath

29 күн бұрын

Thanks for sharing ❤️ =

@Slimmo_09

27 күн бұрын

You need to complete this. Where is final division by 2.

AO = OB is not given. I think that should be clarified.

Thank you!

@PreMath

29 күн бұрын

You are very welcome! Thanks ❤️

My way of solution ▶ Since this is a semicircle, if we consider the triangle ΔABC, then ∠ BCA= 90° OE=ED=DP=PO= a AD= r ⇒ a√2= r a= √2 r/2 ΔABC ~ ΔAPO AP= x OP= a OP= √2 r/2 BC= y CA= 32 ⇒ OP/BC= AO/CA= PA/AB a/y= x/2r= r/32 ⇒ √2 r/2 / y= r/32 y= 16√2 BC= 16√2 BC²+CA²= AB² BC= 16√2 CA= 32 AB= 2r ⇒ (16√2)²+32²= (2r)² 256*2+1024= 4r² 1536= 4r² r²= 384 Ablue= a² = (√2 r/2)² = 2r²/4 = r²/2 Ablue= 384/2 Ablue= 192 square units

@PreMath

29 күн бұрын

Excellent! Thanks for sharing ❤️

Nice problem and solution! I solved it differently, using analytic geometry, but your way was much more elegant. You pronounced the Greek mathematician as "thails" but it should be pronounced "thay-leez". That is two syllables. For confirmation, see kzread.info/dash/bejne/ap53u5OAitDKqJc.html.

@PreMath

29 күн бұрын

Excellent! Thanks for the feedback ❤️

Прошу прощения, но эту задачу я уже решал на этой самой русской олимпиаде у Казакова!

@PreMath

29 күн бұрын

Thanks for the feedback ❤️

STEP-BY-STEP TRIGONOMETRICAL RESOLUTION PROPOSAL : 01) Square Side = X 02) Square Diagonal = X*sqrt(2) 03) Radius of Semicircle = Square Diagonal = X*sqrt(2) 04) Angle (OAP) = a and Angle (OPA) = b. aº + bº = 90º 05) tan(a) = X / X*sqrt(2) ; tan(a) = 1/sqrt(2) ; tan(a) = sqrt(2)/2 06) tan(b) = X*sqrt(2)/X ; tan(b) = sqrt(2) 07) sin^2 (a) = 1/3 and cos^2 (a) = 2/3 08) sin(a) = sqrt(3)/3 and cos(a) = sqrt(6)/3 08) BC / 32 = tan(a) ; BC / 32 = sqrt(2)/2 ; BC = (32*sqrt(2))/2 ; BC = 16*sqrt(2) ; BC ~ 22,63 09) Finding Diameter AB (2R) : 10) AB^2 = AC^2 + BC^2 ; AB^2 = 1.024 + 512 ; AB^2 = 1.536 ; AB = sqrt(1.536) ; AB (2R) = 16*sqrt(6) 11) Radius = 8*sqrt(6) 12) As stated above : Radius of Semicircle = Square Diagonal = X*sqrt(2) 13) R = X*sqrt(2) 14) So : X*sqrt(2) = 8*sqrt(6) 15) X = (8*sqrt(6)) / sqrt(2)) 16) Blue Square Area = X^2 18) X^2 = 64*6 / 2 ; X^2 = 384 / 2 ; X^2 = 192 ANSWER : The Blue Square Area equal to 192 Square Units. Best Regards from The Universal Islamic Institute for Study of Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate.

@PreMath

29 күн бұрын

Excellent! Glad to hear that! Thanks for sharing ❤️