Math Olympiad | Find missing side length m of the triangle

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

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11 ай бұрын

Learn how to find the missing side length m of the triangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Exterior angle theorem; Isosceles Triangles; Congruent Triangles. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Math Olympiad | Find m...
Math Olympiad | Find missing side length m of the triangle | (step-by-step explanation) #math #maths
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Пікірлер: 43

@Fensmiler 11 ай бұрын

Early and i like these videos

@PreMath 11 ай бұрын

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

@Fensmiler 11 ай бұрын

@@PreMathThanks, cheers!

@Istaphobic 11 ай бұрын

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

@richardsullivan1655 11 ай бұрын

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

@egillandersson1780 11 ай бұрын

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 !

@RondoCarletti Ай бұрын

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

@madhusudanaraokuppili1957 11 ай бұрын

Very good explanation professor

@WaiWai-qv4wv 11 ай бұрын

I always respect you❤

@PreMath 11 ай бұрын

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

@wackojacko3962 11 ай бұрын

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 11 ай бұрын

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

@giuseppemalaguti435 11 ай бұрын

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

@soli9mana-soli4953 11 ай бұрын

Nice solution Prof!!

@Abby-hi4sf 11 ай бұрын

Super!

@tsriketwm7274 11 ай бұрын

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

@arnavkange1487 11 ай бұрын

I loved this sum

@PreMath 11 ай бұрын

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

@KAvi_YA666 11 ай бұрын

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

@PreMath 11 ай бұрын

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

@aryanbatra7223 11 ай бұрын

Thankyou Sir! ❤ Happy teacher's day

@bigm383 11 ай бұрын

Thanks Professor!❤

@murdock5537 11 ай бұрын

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

@robertlynch7520 4 ай бұрын

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 ✓ ≡=-⋅

@arnavkange1487 11 ай бұрын

So many constructions u gave but it was understood by me

@PreMath 11 ай бұрын

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

@muphychen7145 10 ай бұрын

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

@pralhadraochavan5179 11 ай бұрын

Good morning sir Happy teachers day

@TheDHemant 10 ай бұрын

@DB-lg5sq 9 ай бұрын

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

@user-ru7pl4tx5v 11 ай бұрын

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.

@Pryszczyk1 11 ай бұрын

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

@albertomarin5264 11 ай бұрын

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

@JSSTyger 11 ай бұрын

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 11 ай бұрын

Wow I just verified that it works :D

@DB-lg5sq 9 ай бұрын

Ne suivez pas la construction

@comdo777 11 ай бұрын

asnwer=3.5cm isit

@DB-lg5sq 11 ай бұрын

But x=?

@kirolosreda7262 11 ай бұрын

Why BAD = ADE😄

@misterenter-iz7rz 11 ай бұрын

tan 4x=5tan x, solving x from it, and m=1/cos 4x.😅

@PreMath 11 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@phlynheubach3745 11 ай бұрын

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