In the first method once found AD, AE and AC as Booster, I've found X with: AE² = AD*AC - DE*CE (4√ 7)² = 6√ 2* 3/2√ 2x - 4*x 112 = 18x - 4x x = 8

The second method is fine, but the first one is incomplete, since it was not demonstrated why AB/AE = BD/ED. The previous step did not correspond to the case of a right triangle with an acute angle bisector, but rather an isosceles triangle. It is easy to prove this missing step, however, by adding a construction line joining D to a point B' on the line AE that is at distance from A equal to the side AB. This gives a triangle DB'E that is similar to ABE, from which the equality AB/AE = BD/ED follows.

Triangle ABD is a 30-60-90 triangle. AB=3√3

I mean, if you're going to use the trigonometric solution, then … why not use the trigonometric solution to figure out the answer without ALL the messy algebra? [1.1]  tan θ = 3 / 𝒂 [1.2]  tan 2θ = (3 ⊕ 4 → 7) / 𝒂 Identities… [2.1]  tan 2θ = (2 tan θ) / (1 - tan² θ) Substitute in left side of [1.2] with [2.1] [3.1]  (2 × 3/𝒂) / (1 - 9/𝒂²) = 7/𝒂 … some algebra gives [3.2]  (6 / ((𝒂² - 9) / 𝒂)) = 7/𝒂 … move stuff around [3.3]  7𝒂² - 63 = 6𝒂² … move more bits [3.4]  𝒂² = 63 [3.5]  𝒂 = 3√7 [3.6]  𝒂 = 7.9373… We can now use that to find (θ) with [4.1]  tan θ = 3 / 𝒂 [4.2]  arctan( tan θ ) = arctan( 3 ÷ 7.9379… ) [4.3]  θ = 20.705…° Cool, now just multiply θ by 3, and take tangent, and multiply by 𝒂, the adjacent: [5.1]  𝒂 tan 3θ = full base [5.2]  7.9373 tan 62.115…° = full base [5.3]  full base = 15 And then just subtract off the 3 and 4 already given [6.1]  𝒙 = 15 - (3 ⊕ 4) [6.2]  𝒙 = 8 That would be the desired (and apparently also solved) solution. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

My corrected solution: My solution uses more geometry, less algebra (no quadratic equation), and no trigonometry. Construction Extend CB toward B to F so BF = BD = 3 Draw AF Solution Triangle ABF is congruent to ABD. So AF = AD, angle BAF = BAD = theta. Use the property of an angle bisector shown on 1:50. AE is the angle bisector of the triangle DAC. --> AC/AD = CE/DE = x/4 AD is the angle bisector of the triangle FAC. --> AC/AF = CD/FD = (x+4)/6 But AD = AF; so x/4 = (x+4)/6 --> x = 8

Thank you for this video. Could you please give me a reference for these Olympiad questions?

1:55 what is the name of the theorem that let's us arrive at those ratios?

I note x = tan(theta), y = tan(2.theta) and z = tan(3.theta). x = BD/AB =3/AB and y = BE/AB = 7/AB, so y = (7/3).x Or y = (2.x)/(1 -x^2), so we have: (7/3).x = (2.x)/(1 -x^2), or 7/3 = 2/(1 -x^2) as x 0 Then 1 - x^2 = 6/7 , x^2 = 1/7 and x = 1/sqrt(7) as x>0. Now we also have 7 = (7/3).x = 7/(3.sqrt(7)) Now z = (x +y)/ (1 -x.y) = (10/3.sqrt(7))/(1 -(1/3)) = 5/sqrt(7). So we have z = 5.x = BC/AB, so BC = 5.BD = 5.3 = 15, and finally EC = 15 - 7 = 8.

Let y be the length of AB. tan θ = 3/y y = 3/tan θ tan 2θ = 7/y y = 7/tan 2θ 3/tan θ = 7/tan 2θ tan 2θ/tan θ = 7/3 2tan θ/(tan θ)(1-tan² θ) = 7/3 2/(1-tan² θ) = 7/3 1 - tan² θ = 6/7 tan² θ = 1 - 6/7 = 1/7 tan θ = √(1/7) = 1/√7 θ = arctan(1/√7) ≈ 20.705° 3/tan θ = (x+7)/tan 3θ x+7 = 3tan 3θ/tan θ x = 3(5/√7)/(1/√7) - 7 = 15 - 7 x = 8

Very beautiful question. Can be easily solved by using trigonometry. But by 1st method , one without knowing trigonometry can also the problem by using Pythagoras theorem & simple geometry. That's the beauty of the problem ❤️❤️

My solution uses more geometry, less algebra (no quadratic equation), and no trigonometry. Construction Extend CB toward B to F so BF = BD = 3 Draw AF Solution triangle ABF is congruent to ABE. So AF = AE, angle BAF = BAD = theta Use the property of an angle bisector shown on 1:50. AE is the angle bisector of the triangle DAC. --> AC/AD = CE/DE = x/4 AB is the angle bisector of the triangle FAC. --> AC/AF = CD/FD = (x+4)/6 But AD = AF; so x/4 = (x+4)/6 --> x = 8

@user-vb8ip1wl5d

3 ай бұрын

correction Triangle ABF is congruent to ABD. So AF = AD, angle BAF = BAD = theta.

@user-vb8ip1wl5d

3 ай бұрын

I have corrected my typo mistakes in another comment.

Please take jee advance questions

Dai teoremi dei seni risultano due equazioni...((7+x)/3)*sinθ/sin3θ=cosθ/cos3θ ..4/x=cos3θ/cosθ..dalle due equazioni e usando le formule sin3x e cos3x...risulta facilmente x=8. .very interesting

X=8, may be