A very tricky interview question: the rectangle in a triangle problem

University entrance exams and job interviews have asked questions similar to this problem. What is the area of the rectangle? What is the minimum area of the large triangle?

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[a.tsuruya]

I'm constantly fascinated by the amount of mental gymnastics that could be avoided simply by using calculus.

[garsox]

I'm glad I have my own business, cause I answered 'tomato' for the second part...

[eccleshillstluke]

For part 2, although using calculus to find the minimum of the area works for those who know calculus, you can actually visualise the minimum. In the big total triangle, cut off the little triangle to the right of the rectangle (which has sides y and 3), and rotate it 180 degrees about its upper-left vertex, and stick it back on so that it looks like a dormer window built onto the roof of the other piece (this is easier to draw than to describe in words!). Then the area of the total is equal to the bottom rectangle plus another identical rectangle above it, plus the remaining bit of the top triangle that sticks above the line. The two rectangles have total area 24, so the total area is 24 plus whatever surplus triangle sticks out. So the way to minimise the total area is to pick the configuration in which the surplus triangle vanishes - namely when the small triangle you rotated exactly matches the top triangle and leaves no surplus. (Maybe one of the 633 other commentators already said this?) Yours JOHN HARTLEY.

[mcmac]

One more thing I like about this channel is that it leaves the viewers the other correct ways of solving the given problems and arriving at the right answer, which is a good learning opportunity.

[TedHopp]

For the first part, if you write the area of the large triangle as (x+3)(y+4)/2 and set it equal to the sum of the areas of the small triangles and the rectangle, 2x+3y/2+xy, then the equation simplifies to xy=12.

[lionelspencer-ward]

Oh... this brings back memories, I remember this coming up when I applied for my first job as a truck
driver, lucky I remembered my calculus.
Questions like this are probably the reason there is a shortage of truck drivers today!

[satrickptar]

When I heard "minimum" I knew I had to do derivatives. I guess it's an instinct as a math student. It takes me back to my optimization lesson.

[abhijiths]

First part was easy. For the second part I tried taking the min of the function 2y+(3/2)x+12 directly. Took a while to notice I needed to use xy=12 and solve it by minimising 2y+18/y+12. In the end finally solved it 😅

[victorregis]

You can use the inequality of means to solve the second part without calculus:
We know that x.y =12 and we want to minimize (4x+3y)/2 +12. We can substitute a=4x and b=3y, giving us a.b=144
We now use that (a+b)/2 >= sqrt(a.b) and we find that (a+b)/2>=12, therefore, the minimum will happen at the equality.
Inputing the value in the area equation gives us 24 as the minimum area.

[geoninja]

I nailed the first part quickly, then equally quickly this was a nice

[markdougherty]

You can solve this by pure geometry. Imagine the hypotenuse of the biggest triangle balancing on the corner of the rectangle. If the two smaller triangles are of different sizes we can rotate the hypotenuse to make the larger triangle smaller and the smaller triangle larger. Thinking in terms of limits it's easy to see that a very small change in angle will make the larger triangle decrease by more than the smaller triangle increases. So the two smaller triangles must be congruent at the minimum. Pick one of those triangles up, flip it around and put it against the other small triangle and we get a bigger rectangle of total area 2xy. So the area of the big triangle is 2x12 = 24. Job done. No algebra or calculus needed.

[AliKhanMaths]

I saw another similar problem on the UK IMC a few years ago, another nice solution! Solving questions like these inspire me to share my own maths tricks, like you do, as well!

[oliverracz]

Funny, I find mathematics, especially geometry very fascinating and solved the first part of the problem in a minute, then got halfway through the second half and got stuck, so I decided to watch the video. From the word "derivative" onward, I understand exactly zero percent of what's being said and done.

[illiil]

area of triangle = 1/2 * (4 + y) * (3 + x) = 12 + 1/2(4x + 3y)
use arithmetic geometric mean inequality on 4x + 3y
you get 4x+3y >= 2 * sqrt(4x* 3y) where equality holds when 4x = 3y
rest is easy

[ishtiakhasan]

2nd part has an easy solution . just think about for which triangle it is going to be minimum . yes, it is isosceles triangle . now in that case the white sub triangles(small triangle) are going to be same to each other and each of their area will be the half of the rectangle(in this case the rectangle is going to be a square).you can think of folding the white part over the blue area. thus you can understand,

"sum of the 2 small(white ) triangle = area of blue square "
so the area of the triangle will be 2*(area of square)=2*12=24.
simple, you don't need calculus

[bettyswunghole]

Oh geez....that's just made me even more terrified of job interviews than I already am...

[ilickcatnip]

We can also minimise the area using
AM ≥ GM
Area(∆) = 4x/2 + 3y/2 + 12,
Let (4x/2 + 3y/2) = k
So area(∆) = k + 12 (1)
Looking at (4x/2 + 3y/2), and by AM ≥ GM, we can say,
k/2 ≥ [(4x/2)(3y/2)]^½
on putting xy=12 and solving we get, k ≥ 12
Putting this in (1), we get
Area(∆) ≥ 24
Therefore, min area(∆) = 24

[vladimirkobarov]

For the part 2 question, we can solve by using inequality of arithmetic and geometric means

[anshulthakur]

1) by similarity
2) we have xy=12
We get factors (1, 12) (12, 1) (4, 3) (3, 4) (6, 2) (2, 6)
Here we get minimum value when x=y or the difference between x and y is minimum so in this case (3, 4) or (4, 3) are if we put (3, 4) we get 24 and if we put (4, 3) we get 24.5

For max we take (1, 12) and (12, 1)
(1, 12)=32
(12, 1)=37.5

Minimum=(3, 4)=24
Maximum=(12, 1)=37.5
:) ✌️

[reneejones]

When you did your calculation you assumed x and y not zero. But x=y=0 also works with the triangle a 3 4 5 right triangle. I think that makes a better "tricky" solution.