Solving An Incredibly Hard Problem From Australia

This problem stumped me! But it has a pretty nice solution. Thanks to all patrons! Special thanks this month to Kyle, Michael Anvari, Richard Ohnemus, Shrihari Puranik.

Source
Emailed to me about 2 years ago, traced to the Australian Intermediate Olympiad given in 2013.

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[MindYourDecisions]

By the way, the official solution listed 3 methods--all fairly similar--and the video is based on method 2 which is where I learned of the rectangle idea. It's fun to see the range of attitudes, from those who found the answer without valid proofs (so I think would earn 0 points on an Olympiad) to those that want a more formal proof (this is from the official solution!). In a YouTube video, like all math and science channels, we strike a balance to engage as many people as possible. For those serious about the math, I suggest to get Olympiad books with real competitions and official solutions. You'll see the standard for proofs is fairly high, and you'll learn many problem solving techniques along the way!

[adamchan]

I thought of it as: b and c are used twice and multiplied together, let's make those as big as possible. Making a = d = 1 leaves 61 to satisfy the sum. And then split that to get 30 and 31

[AntizombiRUS]

If I was asked this question in 15 years I would say "Yes"

[elyakimlev]

It took me less than a minute to realize the values of B and C must be as big as possible, while being as close as possible (30 and 31 or 31 and 30).
That's because of a rule I learned by myself as a kid: dividing a number exactly by 2 and multiplying the results will give the highest value than any other division.
For example, if you have the number 10:
5 * 5 = 25
4 * 6 = 24
3 * 7 = 21
2 * 8 = 16
1 * 9 = 9

If you have the number 61:
30 * 31 > 29 * 32 > 28 * 33.. etc.

[ShaharSigal]

I solved this problem with a different approach:
The expression that we want to maximize has 4 variables and 3 part: ab, bc, cd. From the 4 variables we can see that 2 of them (b and c) repeat themselves more than once in the expression, and their appearance only increasing the total quantity we want to find (adding and multiplying with positive integers).
Therefore, b and c has the largest and equal affect on the maximized expression. Thus we can conclude that to maximize the expression we need to maximize b and c.
To do so we need to minimize a and d, and set them to 1. We now left with b+c=61. As I mentioned, b and c has an equal affect on the expression, so we can choose either one of the options: b=30 and c=31 or b=31 and c=30.

To conclude:
a = d = 1
b = 30, c = 31
Or:
b = 31 and c = 30

[ArchNemesis]

Ah, yeah... I'll "limit" myself to geometry and algebra...

[khanhphaminh]

Mathematic students in high school should follow this channel. Wow. I mean, I'm 26 years old and still get excited with every questions and the way is solved.

[vjekokolic]

Presh: Let's limit our selves to geometry. No calculus. (Than starts to talk about rectangles)
Me: oh no

[joris]

Finally got 1 through logic. since b and c are present 2 times in the equation they have to be the highest possible integer (30 and 31) while a and d need to be the lowest possible integer (1 and 1) which results in 991

[dominickmeisner]

Imagine being the people that have to think up the math problems

[jhoelious]

I, as a 15 yo, would just sit and cry on my desk hahaha but thanks for the video, I definitely learned a lot ^^

[JimPAX]

Just one note on an excellent solution. Once you find the vertex for the “c” parabola, you don’t need to test both integers on either side since the axis of symmetry is at a half-integer. You know that moving + or - .5 from the vertex will result in the same output.

[larsboantonsen]

When you realize that the smallest area you have to subtract is 1. Then the largest rectangle should only be found where height plus length is equal to 63 (h + l = 63).
Which means that the rectangle must look like a square. (h = 32 l = 31 or l = 32 h = 31)
Areas become height times length minus one (h * l-1)
32 * 31 = 992
992-1 = 991

[niec]

This problem is not really that bad at all considering the fact that it was asked in a Maths Olympiad. Replacing d with 63 - a - b- c, you get a quadratic with respect to c. Optimizing that gives you a quadratic with respect to a, maximizing which will give you a = 1 and c = 31. Then you are left with the term ab which is maximized when b is maximized at 30.

[aqibzaman]

I got it in 1 min..

Since b and c are used twice in the below expression they needs to be higher number.. a and b should be minimum suppose a=1 and b=1 remaining 61/2 for b and c..lets suppose 31 for b and 30 for c.
Now substitute them..
1*31+31*30+1x30=991

[burkean]

I started by noticing three things: 1, that b and c are used twice, while a and d are used once. 2, that when multiplying two numbers with a constant sum, the product is highest when the numbers are equal. and 3, that the products have a symmetry so that a and d are indistinguishable, as are c and b.
Notice that points 1 and 2 are in tension: 1 suggests to maximize b and c; 2 suggests to make them equal. So try both.
Maximize the difference gives a, b, c, d = 1, 30, 31, 1 total = 991
keep them equal as possible gives a, b, c, d = 15, 16, 17, 15 total = 767
One other possibility is that the maximum is at some compromise point close to the max solution.
The nearest adjacent point is a, b, c, d = 2, 29, 30, 2 total = 988
Therefore, if the solution space is continuous, (and does not have multiple local maximums) 991 is the maximum.

[utkarshsharma]

Took him two years to solve it.... Kid would be 17 now

[DrZaius]

Easy to find, easy to argue, hard to proof. It's a classical Olympiad problem in that regard.

[schweppesyt]

990 is the answer if you also think
a, b, c, d represent different numbers:
1, 29, 31, 2

[samuelking]

This is the first problem I’ve ever gotten right on your channel, and I got it in less than thirty seconds. I didn’t prove it, I just know that the maximum area for a given perimeter is a square, so I made the biggest and closest thing to a square that I could after taking the smallest allowed values for a and d