Math Olympiad | Find missing side length m of the triangle | (step-by-step explanation)

[@Fensmiler]

Early and i like these videos

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@Fensmiler]

​@@PreMathThanks, cheers!

[@Abby-hi4sf]

Super!

[@Monsieur_Cauchemar-YT]

Did it exactly the same way, but instead of Pythagoras, I used the cosine rule on △DEF, such that: (4 - m)² = m² + m² - 2.m.m.cos(180 - 4x) ⇒ (4 - m)² = 2m² - 2m²[- cos(4x)] ⇒ (4 - m)² = 2m²[1 + cos(4x)] ⇒ 1 + cos(4x) = (4 - m)²/(2m²) ⇒ cos(4x) = [(4 - m)² - 2m²]/(2m²) Now, in △BCD, cos(4x) = 1/m, so: 1/m = [(4 - m)² - 2m²]/(2m²) ⇒ 2m = (4 - m)² - 2m² ⇒ 2m = 16 - 8m + m² - 2m²; and you obtain the same equation: m² + 10m - 16 = 0

[@madhusudanaraokuppili1957]

Very good explanation professor

[@soli9mana-soli4953]

Nice solution Prof!!

[@richardsullivan1655]

I wouldn’t have ever solved this one. Thank you Professor

[@WaiWai-qv4wv]

I always respect you❤

[@PreMath]

Thank you! Cheers! 😀 You are the best, dear ❤️ Keep it up 👍

[@bigm383]

Thanks Professor!❤

[@aryanbatra7223]

Thankyou Sir! ❤ Happy teacher's day

[@KAvi_YA666]

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

[@PreMath]

You are very welcome! Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@egillandersson1780]

This construction make the solution easier but is not obvious to find. My solution involves tan (4x) / tan (x) = 5; it works but not so easier !

[@tsriketwm7274]

Nice one. That’s a cleverer idea than using the double angle formulae

[@arnavkange1487]

I loved this sum

[@PreMath]

Excellent! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@giuseppemalaguti435]

m^2=1+(tg4x)^2=1+(5tgx)^2=1+5(13-sqrt164)

[@wackojacko3962]

Adding auxiliary lines like DE and DF makes the problem obvious too solve but is not obvious at all to discover these constructs ...but I'm finding that trying simple basic auxiliary lines like in this problem, and watching your many tutorials more than 50% of the time usually is the way to proceed and solve. 🙂

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@RondoCarletti]

The great and the small triagle are congruent: The angle x = 18°, 4x = 72°. That makes it easy.

[@albertomarin5264]

Excelente y elegante solución. Hice una solución trigonométrica muy larga.

[@pralhadraochavan5179]

Good morning sir Happy teachers day

[@murdock5537]

This is really amazing, many thanks, Sir! You are great! h^2 = 15 - 10m = m^2 - 1 → (m + 5)^2 = 41 → m > 0 → m = √41 - 5

[@arnavkange1487]

So many constructions u gave but it was understood by me

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@muphychen7145]

Maybe another solution m= -41^1/2-5 belongs to another parallel universes as well as i^2=-1?

[@ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μ9μ]

Καλησπέρα σας από Ελλάδα. μία εναλλακτική λύση αντί των δύο Πυθαγορείων θεωρημάτων στο τέλος θα ήταν να εφαρμόζαμε το θεώρημα Stewart στο τρίγωνο CDE με τέμνουσα την CE την DF. Τα m^3 απλοποιούνται και καταλήγουμε στην ίδια δευτεροβάθμια εξίσωση. Ευχαριστώ.

[@robertlynch7520]

You know … sometimes it is better (really) to work both symbolically, and with trigonometry. Well, at least that's what I think. [1.1]  𝒉 = 5 × tan 𝒙 … height, and also [1.2]  𝒉 = 1 × tan 4𝒙 But, what is (tan 4θ) anyway? well, in a sense it is (tan 2×2×θ), so that's not too hard. Let's use 𝒕 to be (tan θ): [2.1]  tan θ = 𝒕 [2.2]  tan 2θ = 2𝒕 / (1 - 𝒕²) [2.3]  tan 4θ = 2(2𝒕 / (1 - 𝒕²)) / (1 - ((2𝒕)/(1 - 𝒕²))² ) Its not very obvious on the surface, but with some algebraic reduction I got: [2.4]  tan 4θ = 4𝒕(1 - 𝒕²) / [(1 - 𝒕²)² - (2𝒕)²] Which further reduces to [2.5]  tan 4θ = (4𝒕 - 4𝒕³) / (𝒕⁴ - 6𝒕² ⊕ 1) Since we've already established [1.1] and [1.2], then make 'em equal: [3.1]  5𝒕 = (…) … substitute in [2.5], and divide by 𝒕, cross multiply [3.2]  5𝒕⁴ - 30𝒕² ⊕ 5 = 4 - 4𝒕² and shift around [3.3]  5𝒕⁴ - 26𝒕² ⊕ 1 = 0 … which is kind of quadratic, so [4.1]  𝒕² = [5.161250 or 0.038750] … by quadratic solution. √() to 𝒕 [4.2]  𝒕 = [2.27184 or 0.19685] Well, clearly 2.27 is too large (think how tall the (5 × 2.27) bit would be). So, smaller [5.1]  𝒉 = 5 tan θ = 5𝒕 = 0.984251 … so by pythagoras [5.2]  𝒎 = √(𝒉² + 1²) [5.3]  𝒎 = 1.40312 Which (kind of surprisingly) is exactly the same as your answer. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@Pryszczyk1]

ΔACD and ΔBCD similar, so .m/AC = BC/m. m/6 = 1/m, so m= sqrt(6). :)

[@DB-lg5sq]

Deux solutions m^2 égal à 66+ou-10racine41

[@JSSTyger]

Well I came up with this amazingly complicated equation. m^5-m^4-152m³-48m²+776m-576 = 0. How I got there....I used 25+h² = r², 1+h² = m², and cos(4x) = 1-200h²/r^4

[@JSSTyger]

Wow I just verified that it works :D

[@TheDHemant]

5

[@DB-lg5sq]

But x=?

[@DB-lg5sq]

Ne suivez pas la construction

[@축복이-x6u]

asnwer=3.5cm isit

[@kirolosreda7262]

Why BAD = ADE😄

[@phlynheubach3745]

Nice solution. I solved it just using the given two right triangles, using tan(x) and tan(4x) and the angle sum formula for tan(a+b). Found the value of x=arctan(+/-sqrt((26+/-sqrt(656))/10))=11.13634131 degrees (only solution that works). Then used m=1/cos(4x)=1.403124238