Olympiad Mathematics | Find missing side lengths in the triangle | (Fast explanation)

[@mohanramachandran4550]

Fantastic sum Excellent solution

[@PreMath]

Many thanks You are awesome. Keep it up 👍

[@williamwingo4740]

Here's a trigonometric solution: Let AC = b as you did. Then tan x = 9/b and tan 2x = b/12. But tan 2x = (2 tan x)/(1 - tan^2(x)); so b/12 = (18/b)/(1 - 81/b^2) = (18/b)/((b^2 - 81)/b^2); invert the divisor and multiply: b/12 = (18/b)((b^2)/(b^2 - 81)) = (18 b^2)/(b)(b^2 - 81); factor out b on the right: b/12 = (18 b)/(b^2 - 81); cross-multiply: (b)(b^2 - 81) = (12)(18)(b); factor out b again and collect terms on both sides: b^2 - 81 = 216; add 81 to both sides: b^2 = 216 + 81 = 297 = (9)(33); so b = √((9)(33)) = 3√33, which agrees with your answer. From there we proceed as you did, invoking Pythagoras twice to get BC = 21 and AC = 3√42. Cheers! 🤠

[@luigipirandello5919]

Nice. Thank you.

[@ALIabyazu]

It was great

[@harikatragadda]

A circle passes through A and C such that B is the center since AC as a chord subtends angle X in its reflection and 2 X at the center B. Hence the radius of this circle is BC = 9+12 = 21. Now its easy to calculate using Similarity of ∆ADC, ∆ACE and ∆CDE CD² = AD*DE= 9*33 CD = 3√33 AC/AD = AE/AC AC² = 9*42 AC= 3√42

[@Nature-Melody2106]

💥Nice solution💖💖💖

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!!!!!!!

[@noamrtd-g4f]

You can also say the following: DC = 12Tan(2x) & DC = 9/Tan(x) By using the Double angle formula we can get that Tan(2x) = (2tan(x))/(1-tan^2x) So we can say that : DC = 12*(2tan(x))/(1-tan^2x) & DC = 9/Tan(x) Thus: 12*(2tan(x))/(1-tan^2x) = 9/Tan(x) From here by solving, we can get that tan(x) = sqrt(3/11) => x = 27.57 From here it's just simple trigonometry to calculate DC, and then we can use Pythagoream theorem for AC & BC Amazing videos! Keep it up! ❤

[@tombufford136]

I solved this in a similar method, though it took me quite some time.Thank you again.

[@Mycroft616]

And then there is me who used the tan(a + b) identity.

[@petermerrick6615]

Me too, but only if you have done that level of maths. Btw, he could make the Pythagorean calculations easier by scaling down and then back up, which is especially useful in non-calculator scenarios.

[@orobosaisokpunwu6774]

I used the tan identity as well and solved for CD directly

[@Copernicusfreud]

Yay! I solved the problem.

[@bekaluu1]

Used double angle identity for tan to calculate the height of the given triangle. The rest is easy (Pythagoras theorem)

[@ybodoN]

Like two weeks ago, the height (CD) of such a triangle is √(AD² + 2 ⋅ AD ⋅ DB) And here are the new formulas: AC = √(2 ⋅ (AD² + AD ⋅ DB)) and BC = AD + DB

[@marioalb9726]

tan x = 9 / h h = 9 / tan x tan 2x = h / 12 h = 12 tan 2x Equalling: 9 / tan x = 12 tan 2x tan 2x . tan x = 3/4 Clearing Angle: x : 27,575° h = 9 / tan x h = 17,234 cm a = h / cos x a = 19,44 cm b = 12 / cos 2x b = 21 cm ( Solved √ )

[@wackojacko3962]

Once again the pendulum or dance of Pythagorean Theorem. 😉

[@murdock5537]

Nice! DCA = φ; DBC = 2φ; AD = 9; BD = 12; CD = h → tan⁡(2φ) = h/12 → tan⁡(φ) = 9/h → tan⁡(2φ) = 2tan⁡(φ)/(1 - tan^2(φ)) = h/12 = 18h/(h^2 - 81) → h^2 = 297 = 33(9) → h = 3√33 → AC = √(297 + 81) = 3√42 → BC = √(297 + 144) = 21 or: CAB = BCA = 90° - φ → AB = 21 → AD = 9 → BD = 12 → CD = h → AC = b = AE + CE ↔AE = CE = AC/2 → BC = a = 21 → sin⁡(φ) = (b/2)/21= 9/b → b = 3√42...

[@quigonkenny]

As ∠ADC = 90° and ∠DCA = x, ∠CAD = 90°-x. Similarly, as ∠CDB = 90° and ∠DBC = 2x, ∠BCD = 90°-2x. As ∠DCA = x and ∠BCD = 90°-2x, ∠BCA = 90°-2x + x = 90°-x. As ∠BCA and ∠CAB are congruent, ∆ABC is isosceles and AB = BC = 12+9 = 21 ---- (sol 1) Triangle ∆CDB: a² + b² = c² 12² + CD² = 21² CD² = 441 - 144 = 297 CD = √297 = 3√33 ---- (sol 2) Triangle ∆ADC: c² = a² + b² CA² = 9² + (3√33)² = 81 + 297 = 378 CA = √378 = 3√42 ---- (sol 3)

[@gelbkehlchen]

Solution: angle ACB = angle ACD + angle DCB = x+90°-2x = 90°-x angle BAC = angle DAC = 90°-x ⟹ angle ACB = angle BAC = 90°-x ⟹ BC = AB = 9+12 = 21 [because ABC is an isosceles triangle] Pythagoras in triangle DBC: DC = √(BC²-BD²) = √(21²-12²) = √297 = 3*√33 ≈ 17,2337 Pythagoras in triangle ADC: AC = √(AD²+DC²) = √(9²+297) = √378 = 6*√10,5 ≈ 19,4422

[@michaelkouzmin281]

Hi! by the way to figure out all unknowns x = atan(9/(3*sqrt*33)) = D48.13

[@giuseppemalaguti435]

Sine rule.. 21/sin(x+(180-90-2x))=AC/sin2x.. con AC=9/sinx...risukta dai calcoli sinx=sqrt(9/42)..successivamente calcolo tutti i lati

[@JSSTyger]

CB=21 and AC = 3sqrt(42)

[@AnonimityAssured]

This is made more manageable with a little trick. Spoiler alert. First, divide the given lengths by three, as they have it as a common factor, then multiply by three at the end.

[@muslim19800]

🔼 *I like it* 🔼

[@mathexpress9171]

I solve it by trigonometry

[@bigm383]

❤🥂😀📐

[@PreMath]

Thanks for your continued love and support! You are awesome. Keep it up 👍 Stay blessed 😀

[@marioalb9726]

Everytime we have this configuration, with angle x and 2x as figure shows, we have an isosceles triangle, of equal sides 's' = 9+12 = 21 So, side CB = side AB s = 21 cm cos 2x = 12 / 21 x = 27,575° tan 2x = h / 12 h = 17,234 cm tan x = 9 / c c = 17,233 cm ( Solved √ )

[@harrymatabal8448]

English is not a precise language. If the side is missing how can I find it length. First I must find the side. Don't you agree😂