Three tricky math problems.

[barren tusk]

If your age is less than 16, this probably isnt for you.
If not, I challenge you to solve these problems

@bright spruce @hidden sun

[hidden sun]

Is answer of 1st one 117?

[barren tusk]

Wrong. Whats your approach. Also spoiler you ans :)

[hidden sun]

Is it close atleast

👉 👈

[barren tusk]

No

[hidden sun]

Trying again

[minor dawn]

||s = sqrt(3)
x^2 + y^2 ≤ (x-1)^2 + y^2 + (x-1/2)^2 + (y-s/2)^2
0 ≤ xx - 3x + yy - ys + 2
1 ≤ (x-3/2)^2 + (y - s/2)^2
disk radius 1 through B,C
(π/6 - sqrt(3)/4)/(sqrt(3)/4) = 2π/3sqrt(3) - 1
2 - 2π/3sqrt(3)||

[barren tusk]

Nice Problem 3 is done.

[minor dawn]

||2: assume (x,y,z) has length 1, cos(angle) = (x,y,z) * (1,1,1) / sqrt(3)
((x+y+z)^2 + (-x+y+z)^2 + (x-y+z)^2 + (x+y-z)^2)/3
(4x^2 + 4y^2 + 4z^2)/3
4/3
?
||

these are sines mb gang

||so just 4 - 4/3 which is 8/3||

[hidden sun]

I understood only 1 question that too done wrong

[minor dawn]

||3 points would define a unique plane, except...
(1,2,3) doesn't determine a plane,
(2,3,n), (1,3,n), (1,2,n) for n ≥ 4 all determine the same plane, and:
(4,5,6), (3,5,6), (3,4,6), (3,4,5) all determine the same plane. so, we lose one plane, and for each n ≥ 4 we lose two planes, and then we lose three planes, for a total loss of 1 + 14 + 3 = 18
so the answer is 102?||

[barren tusk]

Q1 is the hardest btw. Try other two

[minor dawn]

judging by the number of lines in the question, 1 is harder than 2 and 3 combined

[barren tusk]

Seems like a shorter approach than mine.
I will try to understand it later

[hidden sun]

Oh wait i did read q1 wrong

[barren tusk]

What did u miss

[hidden sun]

Is answer 45

[hidden sun]

P3 was common collinear as well as coplanar.
I took p1p2p3 collinear and p4p5p6p7 coplanar

[bright spruce]

@barren tusk I tried Q3 and I got (2/sqrt3)-1

[barren tusk]

Wring again. Tell your apprach.

[barren tusk]

Wrong

[bright spruce]

I tried again, I got this

,texsp $\frac{2(3 \sqrt 3 - \pi)}{3 \sqrt 3} \approx 0.790800423844$

[barren tusk]

Answers:
Q1. ||102, ||
Q2.|| 8/3||
Q3. ||79% ||

[bright spruce]

:)

[stable lantern]

Banana_314

[barren tusk]

Well done 👍

[barren tusk]

I spoilt the other 2 answers unnecessarily ig

Too bad

[hidden sun]

What level of question is that? Olympiad? Which class?

[barren tusk]

All problems are created by me.
Level is class 12.

[hidden sun]

Im reading this

[barren tusk]

Atleast syllabus wise.
But these problems are unconventional so u may say its somewhat like Olympiad

[hidden sun]

I also generated a self question 2 days b4
I want u to solve it since u fked my mind🐸

Imagine a tanker of height h which is empty initially is being filled with water, water coming out from an oil pipe at the rate of x unit volume/sec. Suppose, initially water is coming alone but oil also began to come from same pipe(mixed with water) such that its rate increases by 2% unit volume/sec, then at what height will the amount of water and amount of oil in the tanker will become equal?
Assume tanker to be cylindrical with radius r

Hope u didn't saw the solution in the thread

Wait there's an error lemme correct

Ok done

[barren tusk]

Never

[hidden sun]

U took 7 min to write this solution

[barren tusk]

Oil increases the flow by 2% so its coming only in very small qty. How can it be equal to the amt of water. Duh

[barren tusk]

No. Not this

[hidden sun]

It's rate is increasing continuously

Read

Also pipe is rigid so if oil flow increases by 2% then by default water flow rate decreases by 2%

[barren tusk]

Wym by 2% unit volume/sec?
Percenage of what? Get ur units right

[hidden sun]

Lemme write question again 😭

[barren tusk]

Ok

[hidden sun]

An empty cylindrical tanker of radius r and is being filled with water coming out from a flow pipe at a rate of $x m^3/s$. Suppose, the flowpipe is rigid so that total flow of volume of fluid at any time is constant. At time t=0,a mixture of oil and water began to flow together from the pipe such that the rate of flow of oil was 0 initially and began to increase at a rate of 2 percent of x per second. At what height will the volume of oil in the taker become equal to volume of water in the tanker?

@barren tusk

Sry there were customers so i took time

[stable lantern]

Ravi Eater

[barren tusk]

Customers?

[barren tusk]

Ok. Rate of change is 0.02 x unit/s² right?

[hidden sun]

🤓

[hidden sun]

Yeah but I ain't telling u step by step

🐸

[barren tusk]

50x/πr²

[hidden sun]

From where u saw the solution

[barren tusk]

Solved

[hidden sun]

It took me 3 helpers to solve🐸

And like 30min

Tho at that time question was itself incomplete

[barren tusk]

Solved it mentally.
Easy to visualize a graph

[barren tusk]

@hidden sun

Area under the graph gives total flow qty

[hidden sun]

U made equation too ig

Coz for time=50 u need them

98.7 wow great guy

[barren tusk]

Wbt you.

[hidden sun]

84 😭

Waiting for april result

[bright spruce]

got the first one ||(102)|| too

It's the same way zfnQRZJT got it btw

[barren tusk]

Great job

[bright spruce]

@barren tusk Got the second one ||(8/3)|| too

||Took a normal to the plane at the centre of the cube (origin).
Took it's coords as (n1, n2, n3) and dotted it with the vectors to the vertices and squared.
Turns out, it's constant and is equal to 8/3||