Intuition is Essential to Problem Solving

#Math #Algebra #inequality

In this video we solve a problem modified from one in the Singaporean Maths Olympiad.

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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[Jop_pop]

So here's a completely different intuition/proof for the problem: you're looking to find the a point in the intersection between a hyperplane of dimension 4 with the 4-sphere of radius 4. The intersection is therefore a 3-sphere. It has a center where all 5 coordinates are equal to 8/5, because the center of the 3-sphere must not move when you swap two coordinates. (This is because both the 4-sphere and the 4-plane are unchanged under coordinate swaps, so their intersection must be as well). To find a point on the 3-sphere, you just start at the center C = 8/5*(1, 1, 1, 1, 1) and add a vector perpendicular to C, since the hyperplane lies perpendicular to C. We're looking to add the vector v = (x, y, z, h, k) with the maximum possible e-component (which is k). Since v is perpendicular to (1, 1, 1, 1, 1), k = -(a+b+c+d). Thus v is the unit vector in the direction (-1, -1, -1, -1, 4). The e-component of that unit vector v is 4/sqrt(20).
Finally, we just need to find how far to go in the v direction, starting from C. We want the unique point P = C + rv, where r is a constant such that P lies on the aforementioned 3-sphere. But there is a right triangle formed by the origin O, point P, and point C. OC is perpendicular to PC because PC lies in the hyperplane, which is perpendicular to OC. Thus PC = r = sqrt(OP^2 - OC^2) = sqrt[16 - 64/5] = 4/sqrt(5).
Putting it all together: P = C + rv. We know r, as well as the e-components of C and v. Those are 4/sqrt(20), 8/5, and 4/sqrt(5) respectively. Plug in to get the e-component of P and simplify. You get P_e = 8/5 + 4/sqrt(20)*4/sqrt(5) = 16/5.

[alexgarcia]

The solution can also be obtained by computing (a-6/5)^2+(b-6/5)^2+(c-6/5)^2+(d-6/5)^2+(e-6/5)^2, which results to be 4, so (e-6/5)^2 <= 4, so e is lower or equal 16/5.

[olsonakin]

Quadratic mean >= arithmetic mean
From there, 5e²-16e<=0
A lot easier than Cauchy-Schwarz inequality

[victorfunnyman]

Wait a minute...

Since a², b², c², d² and e² are all positive numbers (and/or 0)
so to find the maximum, just set a², b², c², d² to 0
set e² to 16
so e = ✓16'
=4
> 16/5 (3.2)
then, with the first equation, you can just put that back in, the other numbers would just have to be 2 or at least average to 1.

max(e) = 4

wait
wait another bit
can't set a, b, c and d to both 0 and 2, that doesn't work
alright then

[남윤지용]

I think we need to explain why max(e)=16/5. Because 0<=e<=16/5 this inequality is necessary condition.

[advaykumar]

You can just use root mean square am inequality

[steffensolgren]

Is this not just a special solution? The more general one you have to consider a, b, c, d being any number OF which the sum OF their squares is less than 16-e^2 and their sum less than 8-e?

[ericzhu]

This problem is from USAMO 1978 problem 1

[faridbouakline]

Dumb question: why didn't you consider the case a =b =c = 0?
The reason I ask is this problem can be easily visualise in term of maximizing the area of the square e within the ( a+b+c+d+e) square?

[deeznutmaster]

I have just had this problem in my school and i couldn't find the answer, thank you so much
for the solution