A Very Nice Math Olympiad Geometry Challenge | 2 Different Methods
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- Опубліковано 6 жов 2024
- A Very Nice Math Olympiad Geometry Challenge | 2 Different Methods
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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.
rule of the bisector
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
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 ✓ ≡=-⋅
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.
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
correction
Triangle ABF is congruent to ABD. So AF = AD, angle BAF = BAD = theta.
I have corrected my typo mistakes in another comment.
Thank you for this video. Could you please give me a reference for these Olympiad questions?
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 ❤️❤️
1:55 what is the name of the theorem that let's us arrive at those ratios?
Please take jee advance questions
Triangle ABD is a 30-60-90 triangle. AB=3√3
Don't you mean ABC? It wouldn't be possible for ABD to be 30-60-90.
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
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
A questão é muito bonita. Eu a fiz de duas maneiras distintas, sendo uma delas igual à sua primeira maneira. Parabéns pela escolha! Brasil Setembro 2024. The question is very beautiful. I did it in two different ways, one of which is the same as your first way. Congratulations on your choice! Brazil September 2024.
X=8, may be