# Solution 92: A (not necessarily) Trisecting Median

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**Published on May 16, 2019**- We find the ratio of two cevians by using a property of symmedian and the Stewart's Theorem. Along the way, we also prove the ratio formula for symmedian.

Congratulations to Thicc Kiwi, JHawk24, scaryowl007, Luis Perez, Bryan C, KRISHNA KAMALAKANNAN, Serengeti Ghasa, Bintang Alam Semesta W.A.M, Mokou Fujiwara, and Nicola C for successfully solving this math challenge question! Thicc Kiwi was the first person to solve the question.

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Prince Ivan23 days agoMy solution:

Let K be on MD s.t. AK bisects

adandapMonth ago^{+1}I'll admit I took an engineer's approach to this problem. I applied the cosine rule to every triangle I could find, then found that I had N equations in N unknowns. Therefore the problem was solved in principle. :)

VeL0ciTYMonth agoWaiting for the physics and Calc ap tests!

VeL0ciTY29 days ago@LetsSolveMathProblems I understand :)

LetsSolveMathProblems29 days ago*Edit: After much debate, I decided to not upload AP FRQ videos for this year, mainly because I believe I cannot post a series of explanation that is of a significantly higher quality than those already posted. If you are interested, turksvids has uploaded all solutions to AP Calculus AB/BC FRQ's from this year: usclip.net/user/turksvids

Gabriel PortoMonth ago^{+4}Great video. I didn't know about Symmedians, thank you so much. Great explanation as always.

Mr5nanMonth ago^{+2}I think I kinda missed something. If you proved that symmedian implies the BD/DC=(AB/AC)^2, don't you have to provide the proof of the other implication as well so that you can say that the angle DAC is theta?

Mr5nanMonth ago@LetsSolveMathProblems yeah that would complete the proof 😊

LetsSolveMathProblemsMonth ago^{+5}Since the foot of the symmedian must be on BC and BD/DC strictly increases as D moves from B to C, it follows that the converse holds as well. In retrospect, I probably should have noted this in the video. I apologize for the possible confusion.

Practical MediocrityMonth ago14:20 you can obtain the result by just considering the large triangle (even though this is not the wisest thing to do).

Alberto ZordanMonth agoAt 1:37 you say for the first of many times in this video a word I can't understand. What is it? "Sedian"? I'm feeling so dumb not being able to grasp this word, maybe I'm not even familiar with what it represents for a triangle. AM is a median to me, whereas AD is not.

Thanks in advance.

Alberto ZordanMonth ago@Benjamin Wang Thanks man!

Alberto ZordanMonth ago@LetsSolveMathProblems I was not even aware of the existence of the word "cevian", let alone its concept, go figure! :D I've never heard of it before, it's already astonishing to me. Now I easily grasp that word, so don't worry about your pronunciation: although it may not be that of a native, it's pretty nice and clear to me :)

LetsSolveMathProblemsMonth ago^{+3}I was saying "cevian," which is defined to be any segment from a vertex of the triangle to the opposite side. I do acknowledge that my pronunciation is not always clear, so please feel free to comment whenever there is a word you are possibly not familiar with. =)

Benjamin WangMonth ago^{+4}Alberto Zordan cevian

moroccan geographerMonth ago^{+1}I am SHOCKED that I was actually able to solve this problem! (I'm 14 by the way)

moroccan geographerMonth ago@LetsSolveMathProblems Thank you so much. By the way I used a completely different method to yours in solving this problem.

LetsSolveMathProblemsMonth ago^{+1}Kudos to you! I would not been able to solve this problem when I was 14. =)

Rohit ChaurasiyaMonth agoHey it's 1:54 AM in my country .... Lemme sleep 😝 ......

TypoMonth ago^{+1}2:25 AM here

Cringe Nae nae babyMonth ago^{+1}Your videos make me wish I was good at math. Much love!

Abdul Alhazred24 days agoそうだね