Math Olympiad | Find missing side length X of the triangle | (Centroid and Medians) |

Learn how to find the missing side length X of the triangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Centroid; Medians; right 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 X of the triangle | (Centroid and Medians) | #math #maths
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
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[@KAvi_YA666]

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

[@PreMath]

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

[@parthtomar6987]

Nice solution sir

[@اممدنحمظ]

تمرين جميل . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

[@じーちゃんねる-v4n]

vector BC=2c BA=2a |c|=9 |a|=12 AE⊥CD ∴(2a-c)･(2c-a)=5a･c-2(|a|^2+|c|^2)=5a･c-2(91+144)=0 ∴a･c=90 ∴x^2=|2a-2c|^2=4(|a|^2-2a･c+|c|^2)=4(144-180+81)=180 ∴x=6√5

[@PreMath]

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

[@Copernicusfreud]

Yay! I solved the problem. I had to review the meaning of a centroid of a triangle on another website. Once I understood what the centroid of a triangle was, the problem was pretty basic.

[@HappyFamilyOnline]

Great explanation👍👍

[@unknownidentity2846]

I call F the point of intersection of the two median lines AE and CD. That means that F is the center of mass of the triangle ABC. The lines AE and CD are divided by F in the ratio 2:1: CF = 2DF AF = 2EF For the right triangles ADF, CEF and ACF we can apply the Pythagorean theorem: AC² = AF² + CF² AD² = AF² + DF² CE² = CF² + EF² x = AC x² = AC² x² = AF² + CF² x² = AF² + DF² + CF² + EF² − DF² − EF² x² = AD² + CE² − DF² − EF² x² = 12² + 9² − (DF² + EF²) x² = 144 + 81 − [(CF/2)² + (AF/2)²] x² = 225 − (CF² + AF²)/4 x² = 225 − x²/4 (5/4)x² = 225 x² = 180 x = 6√5 Best regards from Germany

[@PreMath]

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

[@bigm383]

Thanks Professor!❤😀📐

[@pwmiles56]

Not using the properties of the medians... With F the crossing point, in a change of notation let CF = a EF = b DF = c AF = d By Pythagoras x^2 = a^2 + d^2 [1] 9^2 = a^2 + b^2 [2] 12^2 = c^2 + d(2) [3] [2] + [3] - [1] => 225 - x^2 = b^2 + c^2 Let y = DE so y^2 = b^2 + c^2 = 225 - x^2 ABC and DBE are similar triangles in a ratio of 2, so y = x/2 x^2/4 = 225 - x^2 (5/4)x^2 = 225 x sqrt(5)/2 = 15 x = 2.3.5/sqrt(5) x = 6 sqrt(5)

[@PreMath]

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

[@WernHerr]

My solution: With x:24 = ED:12 you find ED=x/2 Let a, b, c and d be the corresponding sections on the blue lines CD and EA . Then: a²+b²=144, c²+d²=81; together a²+b²+c²+d²=225 a²+d²=x²/4, b²+c²=x²; together a²+b²+c²+d²=5/4*x² So 5/4*x² = 225 or x²=180=36*5 and x=6*sqrt(5)

[@murdock5537]

Many thanks, Sir, this ist great!

[@luigipirandello5919]

Beautiful problem. Thank you, sir.

[@yalchingedikgedik8007]

Thanks PreMath Thanks Sir Geometry exercises are difficult . from my point of view

[@PreMath]

You are very welcome, dear No worries. We are all lifelong learners. That's what makes our life exciting and meaningful! I call these math puzzles "gymnastics for the mind!"

[@jimlocke9320]

At 1:11, the centroid is introduced, but the third median is not shown in the diagram. The viewer is left to assume that the line segment connecting B to the midpoint of AC passes through the intersection of AE and CD. Here's another way to do it. For convenience, rotate ΔABC so AC is horizontal and point B is at the top. Construct DE. Note that ΔBDE and ΔBAC are similar by two pairs of corresponding sides are proportional (length BC is twice BE and length BA is twice BD) and the corresponding angles between them are equal (

[@PreMath]

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

[@АнатолийКривой-ы2и]

CB² + AB² = 5AC²

[@misterenter-iz7rz]

x^2=4a^2+4b^2, 4a^2+b^2=144, 4b^2+a^2=81, so 15a^2=144x4-81, a^2=33, and thus b^2=144-4x33=12, therefore the answer is sqrt(180)=6sqrt(5).

[@PreMath]

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

[@arnavkange1487]

Wow i solved it within 5 minutes

[@PreMath]

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

[@wackojacko3962]

The centroid of B F Sherman's 4 side triangle introduced in 1993 is way Off The Horse! 🙂

[@PreMath]

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

[@marioalb9726]

PreMath intenctionaly makes drawing out of real scale, just for lead us to make mistakes !!!!

[@PreMath]

Not really! However, we must justify our outcome through calculations. You are awesome. Keep it up 👍

[@p.surendra.1876]

You can simply use BPT

[@p.surendra.1876]

Tooooo laaaag , I have to play your video on 2x playback speed

[@ybodoN]

From equations ① and ② we can also calculate that a² = 12 and b² = 33 🤓

[@kaliprasadguru1921]

a²+b² =ED²=45 ED =3√5 X = AC = 2ED = 2x3√5=6√5

[@PreMath]

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

[@kaliprasadguru1921]

Pre Math Sir , Thank you for your compliments . Now I am 79 . Even at this age I have some interest in Geometry . That's why I very often go through your channel . I like your channel very much . Because you explain things lucidly .

[@PreMath]

@@kaliprasadguru1921 Thanks my dear friend. Stay blessed 🙏

[@misterenter-iz7rz]

Very difficult 😮

[@PreMath]

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