Mathematical Olympiad | Find missing side lengths in the triangle | (Easy tutorial) |

[@mathbynisharsir5586]

You are a great teacher sir

[@PreMath]

So nice of you, dear❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@jonchester9033]

The problems you present are as addictive as popcorn. Help! I can't stop.😍

[@HappyFamilyOnline]

Great explanation 👍 Thanks for sharing 😊

[@chrissmith7259]

enjoyed this thanks.

[@PreMath]

Glad you enjoyed it❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@Aryan_1Official]

Pre- eminent explanation sir ! 🥹🥹

[@PreMath]

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

[@ybodoN]

Since DB = 14 we then have AB = 21, AC = x, CB = 2x and cos 120° = −1/2. By the law of cosines: 21² = x² + (2x)² − 2 x (2x) (−1/2) ⇒ x² = 63 ⇒ x = 3√7. To find DC, Stewart's theorem could also be used to solve this special case.

[@PreMath]

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

[@johnbrennan3372]

Nice method

[@devondevon4366]

Answer DC=5.291 AC 7.937 BC 15.874, and DB=14 note 7.927 = 3 sqrt 7 and 15.874 = 6 sqrt 7 and 5.291 = 2 sqrt 7 Using a different method. Let AC= a then BC = 2a Using the law of cosine to find AB c^2 =a^2 + b^2 - 2ab cos 120 degrees = (a)^2 + (2a)^2 - 2(a*2a) -1/2 = a^2 + 4a^2 +2a^2 = 7a^2 c= sqrt 7a Hence AB = sqrt 7a Calculating angle A using the law of cosine a=a , b =2a and c= sqrt 7a a^2= b^2 + c^2 - 2bc cos 'alpha' arccos a = b^2 + c^2 - a^2 ------------------------ 2bc (2a)^2 + (sqrt 7a)^2- (a^2) ------------------------------------------ 2* 2a* sqrt 7a 4a^2 + 7a^2- a^2 - --------------------------- -- 4a *sqt 7a 10a^2 --------- 4a * 2.6457a 10a^2/10.58 a^2 10/10.58 = .94517 Hence, one angle= 19.059 degrees and the other angle= 40.94 degrees ( 180 - ( 120 + 19.059) Let's focus on Triangle ACD with side 7 and degrees 60, 40.94, and 79.06 Using the law of sine to find length AC with c = 7 DC = c * sin 40.94/sin 60.1 = 5.291 AC = c* sin 79.06/sin 60.1 = 7 * sin 79.06/60.1 = 7.937 Hence BC = 7.937 * 2 = 15.874 To find DB Use the Law of Sine 15.874/sine 100.94 degrees = DB/sine 60 degrees 15.874 * sin 60 degrees/sin 100.94 = DB (cross-multiply) 15.874 *0.88294=DB 14 = DB

[@PreMath]

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

[@じーちゃんねる-v4n]

let AC=a then from the angle bisector rule BD=14, from the cosine rule cos120°=(a^2+4a^2-21^2)/(2a^2) ∴a=3√7 dot product of vectors CA･CB=2a^2cos120°=-a^2 ∴ |CD|^2=|(2CA+CB)/3|^2=(4a^2+4CA･CB+4a^2)/9=(4/9)a^2 ∴|CD|=(2/3)(3√7)=2√7

[@harikatragadda]

Let AC = a Extend CD to CF such that CBF forms an Equilateral triangle. ∆DCA is Similar to ∆DFB with sides scaled by 2. Hence, DB = 2*AD = 14 CD = 2a/3 Applying Cosine rule in ∆ACD, a² + (2a/3)² - 2*a*(2a/3)Cos60 = 7² a = 3√7 AC = a = 3√7 CB = 2a = 6√7 CD = 2a/3 = 2√7 Image here kzbin.info/www/bejne/f2fLd4KBgap7pNEsi=HC51ck8iU3RVsdp3

[@soli9mana-soli4953]

You always have nice solution!!!

[@harikatragadda]

@@soli9mana-soli4953 Thank you!

[@alexniklas8777]

From the property of the bisector I determined: BD= 14; AB= 21. Labeled: AC= a; BC= 2a. According to the Pythagorean theorem, AB^2=a^2+(2a)^2-4a×cos(120°), found: AC=a=3√7; BC= 6√7; CD= 2√7 Thanks sir!😊

[@jimlocke9320]

An alternative construction is to extend AC up to the right and drop a perpendicular from B, calling the intersection F.

[@PreMath]

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

[@PreMath]

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

[@JamesDavy2009]

If two products have a common factor, you can apply the algebraic method for the sum or difference of two products and treat the common factor as a variable: 18 × 7 - 14 × 7 = 4 × 7

[@devondevon4366]

Beautiful solution, but below is my approach. Whenever there is a triangle with a 120 degree-angle and one side is twice the other, the degrees are always 120, 40.893, and 19.107 degrees due to the cosine formula : c^2 = a^2 + b^2- 2ab cos C c= sqrt (a^2 + b^2 - 2ab cos C) If let a =a and b=2a c will = sqrt 7a So if a =2, then the sides are 2, 4, and 2 sqrt 7 You will always get a sqrt 7 The reason AB was 21 and not include sqrt 7 was because the sides (as found out later) were 3 sqrt 7 , and 6 sqrt 7 (notice 6sqrt 7 is twice 3 sqrt 7). Hence AB becomes 3 sqrt 7. sqrt 7, but sqrt 7* sqrt 7 = 7, and 7*3 = 21, and that why AB is 21, and DB is 14 (21-7) given that AD is 7 In solving Arccos C ( Using the cosine formula Arccos= b^2 + c^2 - a^2 will always give ________________ 2bc 40.893 degrees and 19.107 degrees for the other two angles From that, the Law of Sine can be used to solve the unknown sides of this triangle.

[@vaggelissmyrniotis2194]

For CD=y using law of cosines on ACD and CBD we will have two quadratic equations with two solutions each, with 2*sqrt(7) be the only common solution therefore y=2*sqrt(7).Similarly using law of sines this time on the above triangles we find the other missing sides

[@misterenter-iz7rz]

First clearly BD=14, let x=AC, 7x^2=21^2, x=AC=3sqrt(7), BC=6sqrt(7), let y=CD, 63+y^2-3sqrt(7)y=49, y^2-3sqrt(7)y+14=0, y=CD=sqrt(7) or 2sqrt(7).😊

[@PreMath]

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

[@misterenter-iz7rz]

​@@PreMathwithout angle bisected theorem, it is hard to discard the wrong case.😢 I attempt to compute CD in two ways, one in ACD, one in BCD, we can conclude 2sqrt(7) is the unique right answer.

[@ankitchoudhary2779]

It could be easier if you apply cosine rule, then you can get direct value

[@timurkodzov718]

I solved this geometry problem and got also BD=14, AC=3*sqrt(7), BC=6*sqrt(7) and CD=2*sqrt(7). I drawed also a triangle with 90°, but inside the triangle ABC. I didn't knew, that s=sqrt(ab-xy) (this is a new knowledge for me). I figured out, that CD=2*sqrt(7), because I applied intercept and pythagorean theorem and figured out, that 2*AC=3*CD.

[@PreMath]

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

[@parthtomar6987]

I gave you 101th like nice solution sir

[@PreMath]

Thanks for liking ❤️ You are awesome. Keep it up 👍

[@devondevon4366]

AC= 7.937 DC=5.291 BD=14 BC=15.874

[@PreMath]

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

[@chelliahRaveedrarajah]

Professor, can you prove angle bisector theorem for us?

[@PreMath]

Nice suggestion! I'll try my best to upload soon. You are awesome. Keep it up 👍

[@howardaltman7212]

Never had seen that formula either. Was able to prove it using the Law of Cosines twice. Is there a better way?@@PreMath

[@wackojacko3962]

@ 9:51 that Cur barking in the background is either a mutt, very unattractive , aggresive , or all three but redeems itself by conCURring with the results of this very cool problem. The dog deserves a treat! 🙂

[@PreMath]

My neighbor's dogs😀

[@phzzxguy]

What an absurdly complicated method to find CD, using an obscure formula that might be interesting to prove, but not good to use. Law of Cosines and some easy trig would take 1/4 of the time.

[@unknownidentity2846]

Let's do it side by side: . .. ... .... ..... Triangle ACD: AC/sin(∠ADC) = AD/sin(∠ACD) AC/sin(∠ADC) = 7/sin(60°) Triangle BCD: BC/sin(∠BDC) = BD/sin(∠BCD) 2*AC/sin(180°−∠ADC) = BD/sin(60°) 2*AC/sin(∠ADC) = BD/sin(60°) [2*AC/sin(∠ADC)] / [AC/sin(∠ADC)] = [BD/sin(60°)] / [7/sin(60°)] 2 = BD / 7 ⇒ BD = 14 Triangle ABC: AB² = AC² + BC² − 2*AC*BC*cos(∠ACB) (AD + BD)² = AC² + (2*AC)² − 2*AC*(2*AC)*cos(120°) (7 + 14)² = AC² + 4*AC² − 4*AC²*(−1/2) 21² = AC² + 4*AC² + 2*AC² 21² = 7*AC² AC² = 21²/7 = 21*3 = 63 ⇒ AC = 3√7 ⇒ BC = 6√7 Triangle ACD: AD² = AC² + CD² − 2*AC*CD*cos(∠ACD) 7² = (3√7)² + CD² − 2*(3√7)*CD*cos(60°) 49 = 63 + CD² − (6√7)*CD*(1/2) CD² − (3√7)CD + 14 = 0 CD = (3/2)√7 ± √(63/4 − 14) = (3/2)√7 ± √(63/4 − 56/4) = (3/2)√7 ± √(7/4) = (3/2)√7 ± (1/2)√7 CD = √7 or CD = 2√7 These two possible solutions must be verified with the triangle BCD: BD² = CD² + BC² − 2*BC*CD*cos(∠BCD) 14² = (√7)² + (6√7)² − 2*√7*6√7*cos(60°) 196 = 7 + 252 − 84*(1/2) 196 = 259 − 42 196 = 217 14² = (2√7)² + (6√7)² − 2*2√7*6√7*cos(60°) 196 = 28 + 252 − 168*(1/2) 196 = 280 − 84 196 = 196 ✓ So CD=2√7 is the correct solution. Summary: AD = 7 BD = 14 AB = 21 AC = 3√7 BC = 6√7 CD = 2√7 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep rocking 👍

[@timurkodzov718]

👍

[@timurkodzov718]

Alternativ: Verlängere die Strecke CD einmal bis zur Länge CB und einmal bis zur Länge AC. Zwei ähnliche gleichseitige Dreiecke => BD=14. Sei M der Mittelpunkt der Strecke BC => DM=7 und AM schneidet CD in einem Punkt S. ASD=90°. Sei O der Mittelpunkt der Strecke BD => AMO=90° nach Umkehrung des Satzes des Thales. CD und MO parallel => nach Strahlensatz MO=(1/2)*CD. SD und MO parallel => nach Strahlensatz SD=(1/2)*MO=(1/4)*CD. Wegen 60° ist CS=(1/2)*AC => nach Pythagoras (AS)²=(AC)²-(1/4)*(AC)² =(3/4)*(AC)². Es gilt auch CS=(3/4)*(CD) =>(3/4)*(CD)=(1/2)*(AC) 3*(CD)=2*(AC) Also: (AS)²=(3/4)*(AC)²=(27/16)*(CD)² Pythagoras anwenden: 49=(27/16)*(CD)²+(1/16)*(CD)² =(28/16)*(CD)² 7=(1/4)*(CD)² => CD=2*sqrt(7) => AC=(3/2)*(CD)=(3/2)*2*sqrt(7) =3*sqrt(7) => BC=2*AC=6*sqrt(7)

[@timurkodzov718]

Ganz kleine Korrektur: ich meinte nicht "wegen Umkehrung des Satzes des Thales", sondern "wegen dem Satz des Thales".

[@giuseppemalaguti435]

😂AC=21,CB=42,CD=sqrt28...non ho controllato i calcoli, ah ah.. Ho fatto tutto col teorema dei seni

[@PreMath]

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