Math Olympiad | A Very Nice Geometry Challenge | 2 Different Methods 

*Outra maneira:* Teorema da bissetriz: x/5=6/AC=> AC=30/x. Por Pitágoras, no ∆ABC: (AC)^2= 6^2 + (x+5)^2 (30/x)^2 = 36 + x^2 + 10x + 25 *(30/x)^2 =x^2 + 10x + 61 (1).* Como x>0 (medida do lado BD), note que nesta última expressão, se estudarmos o divisores de 30, vemos que x=3, satisfaz a equação. A equação (1) é equivalente a x^4 + 10x^3 + 61x^2 -900=0, usando o método de Briot-Ruffini, sendo x=3, uma dessas raízes, temos: x^3 + 13x^2 + 100x +300=0. Nesta última equação, como x>0 então x^3 + 13x^2 + 100x +300>0, isso nos diz que as demais raízes são complexas. Portanto, a única solução real é *x=3.*

I used the angle bisector theorem by doing the following: Let length AC = 𝑦. AB/BD = AC/DC (angle bisector theorem) ⇒ 𝑦/5 = 6/𝑥 ∴ 𝑦 = 30/𝑥 ..... (1) Now, using Pythagoras' Theorem: AC² = AB² + BC² ⇒ 𝑦² = 6² + (𝑥 + 5)² ∴ 𝑦² = (𝑥 + 5)² + 36 ..... (2) Substituting equation (1) into (2): (30/𝑥)² = (𝑥 + 5)² + 36 ⇒ 900/(𝑥²) = (𝑥 + 5)² + 36 ⇒ 900 = 𝑥²(𝑥 + 5)² + 36𝑥² ⇒ 𝑥²(𝑥² + 10𝑥 + 25) + 36𝑥² = 900 ⇒ 𝑥⁴ + 10𝑥³ + 25𝑥² + 36𝑥² = 900 ∴ 𝑥⁴ + 10𝑥³ + 25𝑥² + 36𝑥² - 900 = 0 ..... (3) Now, we have equation (3), which is a quartic polynomial. Looking for an integer solution we find 𝑥 = 3 is a solution, leaving a cubic polynomial with unknown constants as a remainder, such that: 𝑥⁴ + 10𝑥³ + 25𝑥² + 36𝑥² - 900 = 0 → (𝑥 - 3)(𝑥³ + a𝑥² + b𝑥 - 300) = 0 ..... (4) Expanding equation (4), we get: 𝑥⁴ + a𝑥³ + b𝑥² + 300𝑥 - 3𝑥³ - 3a𝑥² - 3b𝑥 - 900 = 0 ⇒ 𝑥⁴ + a𝑥³ + b𝑥² + 300𝑥 - 3𝑥³ - 3a𝑥² - 3b𝑥 - 900 = 0 ⇒ 𝑥⁴ + (a - 3)𝑥³ + (b - 3a)𝑥² + (300 - 3b)𝑥 - 900 = 0 ..... (5) ⇒ 𝑥⁴ + (a - 3)𝑥³ + (b - 3a)𝑥² + (300 - 3b)𝑥 - 900 = 𝑥⁴ + 10𝑥³ + 25𝑥² + 36𝑥² - 900 [equation (5) = equation (3)] Such that we get the following equalities: a - 3 = 10 b - 3a = 61 300 - 3b = 0 So that we get a = 13 and b = 100 ∴ (𝑥 - 3)(𝑥³ + 13𝑥² + 100𝑥 - 300) = 0 Taking the remainder cubic polynomial we look again for an integer solution and find that 𝑥 = -5 (x = -5 is a solution of the cubic 𝑥³ + 13𝑥² + 100𝑥 - 300), and we have a remainder binomial, such that: (𝑥 - 3)(𝑥³ + 13𝑥² + 100𝑥 - 300) = 0 → (𝑥 - 3)(𝑥 + 5)(𝑥² + a𝑥 + 60) = 0 ..... (6) Expanding equation (6), we get: (𝑥² + 2𝑥 - 15)(𝑥² + a𝑥 + 60) = 0 ⇒ 𝑥⁴ + a𝑥³ + 60𝑥² + 2𝑥³ + 2a𝑥² + 120𝑥 - 15𝑥² - 15a𝑥 - 900 = 0 ⇒ 𝑥⁴ + (a + 2)𝑥³ + (45 + 2a)𝑥² + (120 - 15a)𝑥 - 900 = 0 ..... (7) ⇒ 𝑥⁴ + (a + 2)𝑥³ + (45 + 2a)𝑥² + (120 - 15a)𝑥 - 900 = 𝑥⁴ + 10𝑥³ + 25𝑥² + 36𝑥² - 900 [equation (7) = equation (3)] Such that we get the following equalities: a + 2 = 10 45 + 2a = 61 120 - 15a = 0 So that we get a = 8 ∴ (𝑥 - 3)(𝑥 + 5)(𝑥² + 8𝑥 + 60) = 0 Looking at the remainder binomial equation 𝑥² + 8𝑥 + 60 we find Δ = 8² - 4.1.60 = -176, which has complex roots, so, the only real solutions to 𝑥⁴ + 10𝑥³ + 25𝑥² + 36𝑥² - 900 = 0 are 𝑥 = 3 and 𝑥 = -5. 𝑥 = 3 is the only positive solution and as 𝑥 is a measure of distance, the only valid solution is: 𝑥 = 3

Can anyone tell me, what if instead of 5 and 6, we have a and b?

Using angle bisector theorem we may write 6/AC= x /5 ACx =30 AC^2* x^2=900 [AC ^2= 36+25+10x +x^2 =x^2+10x +61] x ^4 +10x^3+61x^2-900=0 Solving this we get two real and two imaginary solutions. One of the real solutions is negative The other is 3 Hence x =3 Comment please

If angle BAC consists of two identical angles, then for calculations we can assume that angle BAC is 2*30 degrees, i.e. angle BAC equals 60 degrees. Therefore angle ACB will be 30 degrees. Now we know that side AC is 2 times larger than side AB because sin(30)=1/2. Therefore we will get such an equation. 6^2 + ( x+5)^2 = 12^2 36 + (x+5)^2= 144 => ( x+5)^2 =144-36=108 x=V108 -5 = 5.3923...

In the drawing we will see that angle BAC consists of two equal angles. To cope with this task, let's assume that angle BAC is 2*30=60, from which it follows that angle ACB must be 30 degrees. So we have a right triangle with angles of 90, 30 and 60 degrees. We can now use the formula that a/b=m where a and b are the sides of the triangle at an angle of 90 degrees. This means that our m is a constant, so this formula works for all triangles with angles of 90, 60 and 30 degrees. For the ratio of the longer side to the shorter side, the constant is V3, which is the square root of 3. Now let's solve this task. BC/AB = V3 => (x+5)/6 = V3 => x+5 = 6*V3 => x =6*V3-5 x=10.3923...- 5 = 5.3923.. .

(6H)/ASino°+ (5H)/ASino° ={ 36H/ASino°+25H/ASino° }= = 61HA/A^2Sino°^2 180°61HA/A^2Sino°^2 = 1.119HA/A^2Sino°^2 1^1.10^103^2 2^52^53^2 1^11^13^2 3^2 (HA/A^2Sino°^2 ➖ 3HA/A^2Sino°^2+2)

Eso me pareció muy fácil para una olimpiada ,sabía cómo resolverlo con solo verlo,por trigonometría 🎉

I think that the two methods were similar to another problem that you did previously on your channel.

tan⁡(2θ) = 12x/(36 - x^2 ) = (x + 5)/6 → x1 = 3 x2, x3 = -2(2 ± i√11) → f(x) = (x - 3)(x^2 + 8x + 60) - btw: 3 weeks ago, you made the same thing...🙂

❤❤❤❤❤❤🎉🎉😊😊

x/6=tgθ...(x+5)/6=tg2θ=(x/3)/(1-x^2/36)..(x+5)(1-x^2/36)=2x...(x+5)(36-x^2)=72x..x=3

How did you break down the 3rd degree equation? It can be divided by (x+-n), n submultiple of 180

tan(θ) = x/6 tan(2θ) = (x+5)/6 tan(2θ) = 2tan(θ)/(1-tan²(θ)) (x+5)/6 = 2(x/6)/(1-(x/6)²) 12(x/6) = (x+5)(1-x²/36) 2x = x - x³/36 + 5 - 5x²/36 72x = 36x - x³ + 180 - 5x² x³ + 5x² + 36x - 180 = 0 1 + 5 + 36 - 180 ≠ 0 ❌ x ≠ 1 8 + 20 + 72 - 180 ≠ 0 ❌ x ≠ 2 27 + 45 + 108 - 180 = 0 ✓ x = 3 x² + 8x + 60 x - 3 | x³ + 5x² + 36 - 180 - (x³ - 3x²) 8x² + 36x - (8x² - 24x) 60x - 180 - (60x - 180) (x-3)(x²+8x+60) = 0 x₁ = 3 | x² + 8x + 60 = 0 √b²-4ac = √8²-4(1)60 = √64-240 = √-176 ❌ x₂, x₃ ∈ ℂ x = 3