Poland Math Olympiad | A Very Nice Geometry Problem

Рет қаралды 25,003

Күн бұрын

Пікірлер: 54

@MarieAnne. 2 ай бұрын

Let O = center of circle. OA = OP = OQ = OB = radius = x/2 Let ∠AOP = ∠POQ = θ → ∠BOQ = 180 − 2θ Using law of cosines in △AOP we get: cos θ = [(x/2)²+(x/2)²−3²]/[2(x/2)(x/2)] = (x²−18)/x² Using law of cosines in △BOQ we get: cos(180−2θ) = [(x/2)²+(x/2)²−7²]/[2(x/2)(x/2)] = (x²−98)/x² Now we setup relationship between these angles to get equation in x only cos 2θ = −cos(180−2θ) 2 cos²θ − 1 = −(x²−98)/x² 2 (x²−18)²/x^4 − 1 = (98−x²)/x² 2 (x²−18)² − x^4 = x² (98−x²) 2 (x²−18)² − x^4 = 98x² − x^4 2 (x²−18)² = 98x² (x²−18)² = 49x² (x²−18)² − 49x² = 0 (x²−7x−18) (x²+7x−18) = 0 (x+2) (x−9) (x−2) (x+9) = 0 x = ±2, x = ±9 Since x is diameter of circle, it must be longer than chords → x > 7 *x = 9*

@holyshit922 6 күн бұрын

I would probably solve this problem the same way If we can see central angle and inscribed angle that subtain the same arc (Two such pairs of angles, one to justify that we have right angle,second to justify that ∠QBA is also θ) Then we need only one cosine rule (after that we can use definition of cosine of acute angle)

@ludmilaivanova1603 2 ай бұрын

@ 6:53 connect A and Q and get a right triangle AQM (angle ACB=90 degrees) which is similar to BPM. from that, 6/a= 7+a/3. 18=7a+a^2. a=2. sinse triangle ABC is isosceles D= BM = 9.

@michaeldoerr5810 2 ай бұрын

The answer is 9 units. And it looks like this a good example of knowing what would be the right circle theorem to make use of in order to justify auxiliary lines. I am kind of wondering, how many circle theorems justify auxiliary lines???

@dantallman5345 23 күн бұрын

Method for those of us operating on the left side of the bell curve: (1) Determine that angle ABQ is double of angle ABP. Define point O at midpoint of AB. AB=twice radius, r. Triangles AOP and POQ are congruent, each has two sides equal to radius and third side is 3. Therefore angles AOP and POQ are equal. By central angle we know that angle ABQ is 1/2 angle AOQ and angle AOP is equal to angle ABQ. (2) Solve for r using law of cosines. Call angle AOP=theta. Note for triangle ABQ, cos(theta)=7/(2r). Apply law of cosines to triangle AOP to get 3^2=2r^2(1-cos(theta)) =2r^2(1-7/(2r)) simplifies to 9=r(2r-7). Arrange into std quadratic form and solve.

@soli9mana-soli4953 2 ай бұрын

A fast solution can be drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x. We can find x with Euclid’s theorem on APB right triangle: 3^2=x*(x-7)/2 x^2-7x-18=0 X=9 Very nice problem, very nice the solution by Mathboost❤

@timeonly1401 2 ай бұрын

soli9mana, would you describe in greater (clearer) detail your construction? I dont' understand the "drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x" (1) I know what an 'isoscele trapezoid' is; I don't understand HOW you got it. (2) I tried to look up "Euclid's theorem", but can only find a theorem from number theory, not one that would apply to right triangles... HELP!! Thx. 😊

@soli9mana-soli4953 2 ай бұрын

@ Sure. You only have to draw a little different figure in which the segments whose length is 3 are one from A and the other from B, at the opposite sides of the diameter, simmetrically. Then the segment of 7 on the top, in the middle, parallel to the diameter, being lateral sides at the same height. In this way you see an isosceles trapezoid , with the diameter, and your goal is to find its major base. I hope it’s clearer, can’t say better

@soli9mana-soli4953 2 ай бұрын

@@timeonly1401 1) I simply built the isosceles trapezoid with the recombined segments (chords if you prefer) of the original figure: instead of having the sequence 3-3-7, replace with 3-7-3. Or try to draw an isosceles trapezoid inscribed in the semicircle with the major base x, minor base 7 and the two sides 3, in this way you will have your modified figure. 2) If you are a fan of geometry problems, you should know the two beautiful theorems of Euclid, the man who wrote the most beautiful mathematics book in the world! Know that there are a couple that concern right triangles, the one I used in the problem exposed says that: In any right triangle, the area of the square on a side adjacent to the right angle is equal to the area of the rectangle whose dimensions are the length of the projection of this side on the hypotenuse and the length of the hypotenuse. I Hope this help

@oscarcastaneda5310 2 ай бұрын

Knowing cos(theta) = 7/(2r) and then using the Law of Cosines with one of the triangles with two sides "r" and third side "3" one can determine that 2r is 9. Theta is the angle subtended by the "3" side and the center of the semicircle.

@kateknowles8055 2 ай бұрын

Adding -{lines AQ and}- BP to the diagram and labelling O the centre of AB ( X is the diameter of the semicircle and O is the centre.) , -{ then APB and AQB are two rightangled triangles with 90º angles at P and at Q respectively. AQ^2= X^2 - 7^2 and BP^2 = X^2 - 3^2 by Pythagoras' theorem. AC.CQ is equal to BC.CP if we label the point of intersection of AQ and BP as C. A, P, Q and B are given points on a (semi-)circle. (AC+CQ)^2 = AQ^2 = AC^2 +CQ^2 +2AC.CQ (BC+CP)^2 = BP^2 = BC^2 + PC^2 + 2BC.CP Now BP^2 - AQ^2 is equal to BC^2 +PC^2 - AC^2 - CQ^2 +(2BC.CP- 2AC.CQ , this term in the bracket equals zero)}- but I think there should be a solution that does not use AQ and does not use C Let ½X = r the radius, which equals OA, OP, OQ and OB Now angles AOP and POQ are equal and naming them each tº then the value of angle QOB is 180º- 2tº and 7x7 = 49 = r^2 +r^2 - 2.r.r cos(180-2t) and 3x3 =9 = r^2 +r^2 - 2.r.r cos (t) 49-9 = 2.r.r ( cos (t) - cos (180 -2t) r.r = r^2= (40/2) / ( cos(t) + cos (2t)) cos (2t) = cos^2(t)-sin^2(t) = 2cos^2(t ) -1 r^2 = 20 / (2cos^2(t) +cos(t) -1) the denominator factorises as ( 2c+1)(c-1) where c= cos(t) So a couple of paths i have shown here which are to be abandonned and the solution to be sought in the video provided here. Peeping at it now I perceive that I forgot that construction can be extended beyond the given shape, and this can simplify the solution. Thank you again, Math Booster , well done to students who " got it" on their own efforts, and encouragement and better "luck" next time to others!

@marcgriselhubert3915 2 ай бұрын

We use an orthonormal center O, the center of the circle, and first axis (OA). We have A(R; 0) B(-R; 0) P(R.cos(t); R.sin(t)) Q(R.cos(2.t); R.sin(2.t)) with R the radius of the circle and t = angleOAP, 0°

@imetroangola17 2 ай бұрын

*_Solução:_* No triângulo retângulo ∆APB: *sen θ = 3/x* No triângulo retângulo AQB: *cos 2θ = 7/x* Como cos 2θ = 1 - 2sen² θ, então: 7/x = 1 - 2 × (3/x)² Seja 1/x = y > 0. Daí, 7y = 1 - 18y² → 18y² + 7y - 1 = 0. ∆ = (-7)² - 4× (18)×(-1) = 121 y = (-7 + 11)/36 = 4/36 = 1/9 x = 1/y → *x = 9 unidades*

@aromaclinic4112 9 күн бұрын

△OAP OA=OP＝r r^2-(7/2)^2=3^2-(r-7/2)^2 r=9/2 x=2r=9

@bpark10001 2 ай бұрын

You can simplify this problem by rearranging the line segments with the 7 long one at the center. Then symmetry gives 2 triangles: one hypotenuse 3, one leg r-3.5, & the other hypotenuse r, one leg 3.5. Equating Pythagoras for the 2 other legs gives equation 3² - (r - 3.5)² = r² - (3.5)². 2r² - 7r - 9 = 0. r = 4.5. X = 2r = 9.

@kateknowles8055 2 ай бұрын

That is a new and useful approach . Well done.

@imetroangola17 2 ай бұрын

*_Solução:_* No triângulo retângulo ∆APQ: *sen θ = 3/x* No triângulo retângulo ∆AQB: *cos 2θ = 7/x* Como cos 2θ = 1 - 2sen² θ, então 7/x = 1 - 2 × (3/x)² Seja 1/x = y > 0. Daí, 7y = 1 - 2 × 9y² → 18y² + 7y - 1 = 0. ∆ = (-7)² - 4×(-1)×18 = 121 y = (-7 + √121)/36 = (-7 + 11)/36 y = 4/36 = 1/9. Portanto, 1/x = y → x = 1/y = 1/(1/9) *x = 9 unidades*

@kateknowles8055 2 ай бұрын

@@imetroangola17 Bueno.

@marcusdecarvalho1354 Ай бұрын

I don't understand how I can simplify that problem.

@bpark10001 Ай бұрын

@@marcusdecarvalho1354 Re-arranging the order of the line segments does not affect the radius, so they can be re-arranged to exploit symmetry to simplify the problem.

@pwmiles56 2 ай бұрын

Taking perpendiculars from the centres of the chords and forming right triangles 2 arcsin(3/x) + arcsin(7/x) = pi/2 2 arcsin(3/x) = pi/2 - arcsin(7/x) Take sine of both sides 2 (3/x) sqrt(1-9/x^2) = sqrt(1 - 49/x^2) 36/x^2 (1-9/x^2) = 1 - 49/x^2 Multiply by x^4, rearrange and simplify x^4 - 85x^2 + 36.9 = 0 x^2 = (85 +/- sqrt(85^2 - 36^2)) / 2 x^2 = (85 +/- 77) / 2 x^2 = 81 or 4 x>=7 so x=9

@kateknowles8055 2 ай бұрын

This looks good , efficient and neat . ( I lose the plot somewhere in line four or five. )

@pwmiles56 2 ай бұрын

@@kateknowles8055 For example, in the third line of working, by the double angle formula for sine sin(2 arcsin(3/x)) = 2 sin(arcsin(3/x)) cos(arcsin(3/x)) The sin and arcsin cancel out so the first term on the right is just 3/x. The second term is cos of something that 3/x is the sin of, so using cos^2 = 1 - sin^2 you get sqrt(1 - 9/x^2). With sin(pi/2 - arcsin(7/x)), you have the complementary angle so it is cos(arcsin(7/x)) and proceed similarly.

@timeonly1401 2 ай бұрын

Draw segments AQ & PB, and call their lengths c & d, respectively. ∠APB & ∠AQB are right angles (By Thales Thm). So, applying Pythagorean Thm to ∆AQB & ∆APB: c² + 49 = x² and 9 + d² = x², so that c=√(x²-49) and d=√(x²-9). All the vertices of quadrilateral APQB are on the semicircle (&, hence, the circle), so APQB is a CYCLIC quadrilateral. So, the product of the diagonals is equal to the sum of the products of opposite sides (Ptolemy's Thm). Applying, we get: (AQ)(PB) = (AP)(QB) + (PQ)(AB) c d = (3)(7) + (3)x = 3(x+7) Subbing in c and d from earlier: √(x²-49) √(x²-9) = 3(x+7) Squaring both sides and solving for x: (x²-49)(x²-9) = 9(x+7)² (x-7)(x+7)(x²-9) = 9(x+7)² [x is a diameter which is the longest chord in the circle, so x > 7. Hence x-7 > 0. And because x-7 is not zero, we can divide both sides by it.] (x-7)(x²-9) = 9(x+7) x³ - 7x² - 9x + 63 = 9x + 63 x³ - 7x² - 18x = 0 x(x² - 7x -18) = 0 [x > 7 ⇒ x ≠ 0] x² - 7x -18 = 0 (x-9)(x+2) = 0 x = 9 or x = -2 (elim., since x>7) so x=9. Done!!

@ekoi1995 Ай бұрын

similar triangles: (a+7)/(6) = (3)/(a) a=2 x=a+7 = 2+7=9

@skwest 2 ай бұрын

Or, instead...... 1) Connect AQ. 2) Drop perpendicular from P to AQ, intersecting AQ at R. 3) Extend PR through O to intersect circle at S, making PS a diameter (length X), with PR = a, and RS = X - a. 4) For Right△APR, calculate (by Pythagoras) the length of AR as √(3² - a²) = √(9 - a²). 5) Use Intersecting Chords (AQ and PS) to figure 'a' in terms of X... i.e. AR * RQ = PR * RS: √(9 - a²) * √(9 - a²) = a * (X - a), or 9 - a² = aX - a², so that 9/X = a 6) Next, use Right △AQB and Pythagoras to establish that: 7² + [2 * √(9 - a²)]² = X², which is, 49 + 36 - 4a² = X²; simplified to 'a' in terms of X becomes: √[(85 - X²)/4] = a 7) Substituting for 'a' from #5, we get: √[(85 - X²)/4] = (9/X), square both sides to get (85 - X²)/4 = 81/X², or 85X² - X⁴ = 324, or X⁴ - 85X² + 324 = 0 8) Substituting Z = X² we have: Z² - 85Z + 324. 9) Use the quadratic equation to solve for Z: Z = [85 ± √(85² - 4 * 1 * 324)]/2, or Z = [85 ± √(7225 - 1296)]/2 Z = [85 ± √(5929)]/2 = [85 ± 77]/2, or Z = 81 or 4 10) Reverting with Z = X² we get: X = √81 = 9, or X = √4 = 2 Obviously, X cannot be equal to 2, as the resulting △AQB would be degenerate. Therefore, X = 9. Q.E.D.

@marioalb9726 2 ай бұрын

My solution, same as video. Excellent !!!

@kateknowles8055 2 ай бұрын

Congratulations and best wishes

@shaozheang5528 2 ай бұрын

The answer is 14. Use the properties of cyclic quadrilaterals and the thales theorem to find out by connecting A to Q.

@raghvendrasingh1289 2 ай бұрын

❤ By Ptolemy's theorem 3x+21 = sqrt {(x^2 - 49)(x^2 - 9) } x^3 - 67x - 126 = 0 X= 9 by RRT

@sarantis40kalaitzis48 Ай бұрын

x^3 -7•x^2 -18x=0, x>0 so x^2-7•x-18=0 so (x+2)•(x-9)=0 so x=-20 Accepted.

@tezchan_p6mk2sr Ай бұрын

円の中心（ABの中点）を点Oとすると △OBP∽△PQA（∵△OBP、△PQAは共に二等辺三角形、∠PQA=∠PBA（弧PAに対する円周角））よってPA:AQ=OB:BP　→PA×BP＝AQ×OB PA＝3、AQ=√(x²-7²)、OB＝x/2、BP＝√(x²-3²) 3×√(x²-3²)=√(x²-7²)× x/2 9x²-81=(x⁴-49x²)/4 x⁴-85x²+324=0　→x²=4、81　→x=2、9 x＞7より、x=9

@AmirgabYT2185 2 ай бұрын

x=9

@gzimatipi3964 Ай бұрын

AQ= √3²+3²=√18=3√2 X=√ (3√2)²+7² X=√9x2+49 X=√67 X=√9*7 X=3√7

@alipourzand6499 2 ай бұрын

Nice solution! I was stucked ☺

@tocamelosuevos 2 ай бұрын

Hay algo que no me cuadra. El problema planteado tiene infinitas soluciones si se resuelve con un compás y una regla, dependiendo del ángulo que arbitrariamente formemos entre los segmentos PA y PQ y, por ende, de la longitud del segmento AQ. Cuando un problema geométrico tiene una 'única' solución trigonométrica, tiene tambíen una 'única' solución gráfica. Y no es el caso.

@giuseppemalaguti435 2 ай бұрын

APQ=α...teorema del coseno 18-18cosα=x^2+49+14xcosα..x^2+31+(14x+18)cosα=0...poi,teorema del seno √(x^2-49)/sinα=3/sin(90-α/2)=3/cosα/2..(x^2-49)/(1-(cosα)^2)=9/(1+cosα)/2….semplifico risulta cosα=(67-x^2)/18..sostituisco mi risulta una cubica 7x^3-469x-882=0,con soluzione x=9

@nexen1041 2 ай бұрын

That was a really good question

@murdock5537 2 ай бұрын

Nice! φ = 30°; AO = BO = PO = QO = r → 2r = AB = x = ? AP = PQ = 3; BQ = 7; QA = k; sin⁡(BQA) = 1 = sin⁡(BPA) BQO = OBQ = AOP = POQ = δ → AOQ = 2δ; ∆ ABQ → k^2 = 4r^2 - 49; cos⁡(δ) = 7/2r ∆ AOP → 9 = 2r^2(1 - cos⁡(δ)) = 2r^2(1 - 7/2r) → (r - 7/4)^2 = (11/4)^2 → r > 0 → r = 9/2 → x = 9 → cos⁡(δ) = 7/9; k = QA = 4√2; QPA = γ → cos⁡(γ) = -7/9 → γ = 6φ - δ; AOQ = 2δ → cos⁡(2δ) = 17/81

@Arroz_040 Ай бұрын

Que pedo pq no hay nadie que hable español xd

@prossvay8744 2 ай бұрын

X=2(9/2)=9

@SGuerra 2 ай бұрын

🎉🎉🎉 A questão é muito boa. Eu resolvi aplicando o Teorema dos Cossenos e a minha solução não exige uma construção auxiliar. Parabéns pela escolha 🎉🎉🎉 BRASIL Novembro de 2024.

@marcelowanderleycorreia8876 2 ай бұрын

Tbm resolvi usando a Lei dos cossenos. 👍

@sarantis40kalaitzis48 Ай бұрын

Applying Ptolemys Theorem we have 3•x+3•7=√(x^2-7^2)•√(x^2-3^2) so [3•(x+7)]^2=(x+7)•(x-7)•(x^2-9) , divide Both Sides by x+7>0, hence 9•(x+7)=(x-7)(x^-9) so 9x+63= x^3-9x-7x^2+63 so x^3-7x^2-18x=0,x>0 so x^2-7x-18=0 so (x+2)•(x-9)=0 so x=-2,x>0 Rejected ,or x=9 Accepted.

@nenetstree914 Ай бұрын

@CristinaMaria-n5p 26 күн бұрын

Bom dia a todos que contam todos os dias com isso trabalho. Tivemos alguns problemas de ordem Matrimonial. Primeira Secretária Segunda Secretária. Mas no mais estamos bem. Entrem em contato comigo.

@CristinaMaria-n5p 26 күн бұрын

Raíz cúbica a Cristina Maria de Souza consegui mostrar e os motoristas que sofreram com descaramento da Maria Nathalia Rufino de Farias Que estavam em outro País. Correram perigo de Morte.