How To Solve 'One Of The Hardest Problems Ever'

This question stumped some of the smartest maths students in Australia, and there's no way I would have solved it in an exam! Can you figure it out?

Syndey Morney Herald and video solution

Official test and solutions available at

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Connect with me

[SatishYadav-xqoy]

Did you figure it out?🤨





Always sounds sarcastic to me

[ryanto]

Did it bother anyone that the final answer for the second part didn't have a simplified radical denominator?

[panchito]

90% of the comments are "Im Indian and I solved this problem under 10 seconds"

[affin]

Wow the comment section is full with intelligent people

[ghazalaanjum]

I'm Indian and couldn't solve it in 2 minutes... Is my nationality safe?

[rishabhmittal]

Quite easy prblm if u knw proportionalty theorem of similar triangle

[blomblorpf]

The first thing I did, after giving credit to the question for requiring a certain amount of intuition and creativity, was to scroll down and look what people have to say, and lo' and behold expectations ons these videos' comment sections are always met. If you found the problem too easy, great, it means you have a good sense for problem solving. If you didn't, great, now you know where you can improve. If you like boast about how easy it is, complain that children stuggling with it is absurd, or if you just like to try and let everyone know how good you are, please use the god-level problem solving skills to solve your egocentric attitude and find a renewed view on your way of going about these situations.

Thank you.

[mdshahnwaz]

I solved...
After watching the video😎

[akashsudhanshu]

I am in 10 standard IN INDIA
I SOLVED IT! THE METHOD INCLUDES CONSTRUCTION,
HONESTLY
TOOK 4 HOURS

[not_vinkami]

6:52 "These videos build confidence for students..."
Wait, how come my confidence on math is getting lower and lower after watching those videos?

[christianfunintuscany]

Not so difficult as I expected from the presentation :)

[tonyhasler]

Great whilst sipping wine at the laptop...less fun in an exam at 17!


I was pleased to get this right after about 45 minutes but I was (incorrectly) convinced after 30 seconds that DZ/DE was 1/sqrt(2) giving ZE/DE as (sqrt(2)-1)/sqrt(2). It just looks that way! I then wasted some time trying to prove this incorrect assumption. Only later did I realise that it was FC/BC that was (sqrt(2)-1)/sqrt(2) and that FC = ZE so I was on the wrong track (or rather my track required adjustment by a factor of sqrt(2) :-)).


I understand that in "big tech" interviews questions like these are posed to see how people respond under pressure but it seems unfair to set a trap like this in a child's exam; unless the child answers 10 or 20 questions like this in one exam it is merely a matter of chance whether the child spots the trap or not.

[jgoemat]

Finally, a fun one that I could (mostly) solve :) The main difference from your solution is that instead of talking about parallelograms, I showed that DZ = YE, so DZ-YZ (which is DY) is the same as YE - YZ (which is ZE)

[guilhermemateus]

The video is called "One Of The Hardest Problems Ever". However, if you read the coments, everyone sims to have got it! And in record timing too!. Wow guys you are all so good. The audiance of this channel has only members with 7000IQ +.

[sanjaygupta]

I learned a lot from this . Thank you so much.😃😃😃

[o_ogggamingo_o]

6:47 You still need to rationalize the denominator, Presh!

[mrbangkockney]

I think this is the first time I’ve ever heard you say you couldn’t have solved under exam conditions so +1 for the honesty.

[brainboxanky]

Presh you are doing a great job by exchanging the maths knowledge from different parts of the world to across the world
This would help in homogenization in mathematics in whole world quite soon

[MikeBTek]

Another way to solve the YZ to BC ratio. We are first given that DE is parallel to BC and BC/DE = √2. This means that all the dimensions of similar triangle ADE to similar triangle ABC have this √2 ratio, including the altitude. If altitude of ABC is say K, then altitude of ADE is K/√2. The difference is K - K/√2 or K(1 - 1/√2), which is the distance between the parallel lines. Every time we draw one of those parallel lines we end up with a similar triangle with altitude reduced another (1 - 1/√2). Since we have drawn 3 parallel lines the resulting triangle in the middle XYZ has an altitude equal to K ( 1 - 3( 1 - 1/√2) ), which computes to K (1 - 3 + 3/√2 ), or K ( -2 + 3/√2 ). This ratio extends through out ABC and XYZ meaning that YZ/BC is (-2 + 3/√2), the same answer obtained in the video.

[futuredystopia]

Well … that was simple. I though you had a hard one for us this time.