Ukrainian Math Olympiad | A Very Nice Geometry Problem

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

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2 ай бұрын

Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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Пікірлер: 23

@joseluishablutzelaceijas928 Ай бұрын

Interesting, thank you for the solutions. Another approach similar to your second one would be as follows. Choose point P on AB s.t. [angle BDP] = 60°, then [DP] = 3 as the triangles DCA and DPA are congruent. Then choose point Q on BD s.t. [angle DQP] = 60°. One than not only has that triangle PQD is equilateral but, in particular, that QP is parallel to DA. Thus the triangles BQP and BDA are similar, therefore [PQ]/[BQ] = [AD]/[BD] = 3/2 = x/5, which leads to x = 15/2 = 7.5.

@RAG981 Ай бұрын

Brilliant. Well done.

@nantesloire Ай бұрын

Sehr schön .

@phungcanhngo 11 күн бұрын

Thank you professor.2nd method is very cool!

@user-kx2ne1in9l 27 күн бұрын

Another solution: draw the bisector of the external angle A and let E be its intersection with base BC. By external angle bisector theorem we get EB/EC=AB/AC or (8+EC)/EC=5/3, thus EC=12. Triangle DAE is right, its hypotenuse is DE=12+3=15 and its angle E is 90-60=30 degrees, therefore AD=15/2.

@murdock5537 Ай бұрын

Nice! φ = 30°; ∆ ABC → AB = AD + BD = 5 + 3; BCD = DCA = θ; CDB = 2φ; DBC = γ; CD = k AC = n; BC = m → n/5 = m/3 → 3n = 5m → n = 5m/3 → AC = 5x → BC = 3x → 15x^2 - 15 = k^2 → k = √(15(x^2 - 1)); sin⁡(2φ) = √3/2 → cos⁡(2φ) = sin⁡(φ) = 1/2 sin⁡(4φ) = sin⁡(6φ - 4φ) = sin⁡(2φ) = √3/2 → cos⁡(4φ) = -cos⁡(6φ - 4φ) = -cos⁡(2φ) = -1/2 ∆ ABC → 64 = 25x^2 + 9x^2 - 30x^2cos⁡(2θ) → cos⁡(2θ) = (17x^2 - 32)/15x^2 ∆ BCD → 9 = k^2 - 9x^2 - 6kxcos(θ) → cos⁡(θ) = 4(x^2 - 1)/x√15(x^2 - 1)) → cos^2(θ) = 16(x^2 - 1)/(15x^2) → sin^2(θ) = 1 - cos^2(θ) = (16 - x^2 )/(15x^2 ) → ∆ BCD → sin⁡(2φ)/3x = sin⁡(θ)/3 → sin⁡(θ) = √3/2x → sin^2(θ) = 3/4x^2 → cos^2(θ) = 1 - sin^2(θ) = (4x^2 - 3)/4x^2 → cos⁡(2θ) = cos^2(θ) - sin^2(θ) = (2x - 3)/2x^2 = (17x^2 - 32)/15x^2 → x = √19/2 → k = 15/2 → 3x = m = 3√19/2 → n = 5√19/2 → cos⁡(θ) = 4√19/19 → θ ≈ 23,41°→ cos⁡(γ) = -√19/38 → γ ≈ 96,59°

@constantinfedorov2307 Ай бұрын

Вообще-то эта задачка решается в лоб без применения интеллекта. В немного более общем виде, пусть отрезки стороны BC m и n (m > n), биссектриса L, тогда существует число x>1, что AB = mx BC = nx; длина биссектрисы L² = mnx² - mn; по теореме косинусов L² + n² - Ln = n²x²; => mnx² - mn + n² - n √(mn)√(x²-1) = n²x²; (m-n)(x²-1) = √(mn)√(x²-1); √(x²-1) = √(mn)/(m-n); L = mn/(m-n); при m = 5 n = 3 L = 15/2;

@DominiqueCathala Ай бұрын

Thank you for all these challenges. A small problem to submit to you: 'rue du chat qui peche' is a narrow street in Paris. Two ladders are placed across the street, one 3 meters, the other 2 meters. They cross at a height of 1 meter. How wide is the street?

@SuperPassek Ай бұрын

Interesting solutions! Another solution is drawing a circumcircle of ABC. Let E be the intersection of the circumcircle of ABC and the extension of AD, and H be the perpendicular foot from E on BC. Then EBC = EAC = EAB = ECB, so EB = EC. It means HB = BC = 4, so HD = 1. From EDH = 60 and HD = 1, we get DE = 2. AD * DE = BD * CD, so AD * 2 = 3 * 5, which yields AD = 15/2

@rabotaakk-nw9nm Ай бұрын

👍, but... HB=HC=BC/2=4 !!!

@SuperPassek Ай бұрын

@@rabotaakk-nw9nm Oh, thank you. Yes, it was a typo. 😅

@FATFATHER506 Ай бұрын

There is a easy way and easy to use 1.from C to AB at E make AE=AC=3y then EB=2y 2.bottom area Tri ECB can be split into 3A and 5A 3.Tri ACE : Tri BCE = 2:3=8A : 12A 4.lenth AD =15/2

@Alejandro52rus Ай бұрын

Very nice methods! But for me it's difficult to see if 2 triangles are similar. So in my solution I used the law of sines twice: 3/sinθ = 2a/sqrt(3) -> sinθ = 3*sqrt(3)/2a and 5/sinθ = 2b/sqrt(3) -> sinθ = 5*sqrt(3)/2b, where a and b are unknown sides of the triangle. So 3*sqrt(3)/2a = 5*sqrt(3)/2b -> b = 5a/3. Then I used the law of cosines for the bottom side of the triangle: 64 = 25a^2 / 9 + a^2 - 2*(5a^2 / 3)*cos2θ. cos2θ = 1 - 2*(sinθ)^2 = 1 - 27/2a^2, then I found a: a= 3*sqrt(19)/2 and used cosines law again for a: (3*sqrt(19)/2)^2 = x^2 + 9 - 2x*3*sin60, solved it and that gave me the same answer

@RAG981 Ай бұрын

Method 2 is lovely.

@jimlocke9320 Ай бұрын

Another approach would be to drop a perpendicular from A to BC and label the intersection as point E. We note that ΔADE is a special 30°-60-90° right triangle. Let DE = a, then AE = a√3 and AD = 2a. Then, CE = 3 - a, tan(Θ - 30°) = (3-a)/(a√3) and tan(Θ + 30°) = (a + 5)/(a√3). Using the tangent sum of angles formula, tan(α + ß) = (tan(α) + tan(ß))/(1 - tan(α)tan(ß)), twice, noting that tan(30²) = 1/√3 and tan(-30²) = -1/√3, we get two expressions for tan(Θ) which can be set equal to each other and solved for a, which is doubled to find AD. This method will work if Θ < 30°, in which case a will be greater than 3, which is the case here (a = 3.75). Point C is to the left of point E. The diagram is misleading because point E appears that it would be between C and D.

@nantesloire Ай бұрын

Die Größe BD = 5 spielt hier überhaupt keine Rolle und wird nicht verwendet ?