Poll of the Day > Only HALF of you will get this MATH PROBLEM Right!! Can you guess it???

(+) Yeah! (+)

A Tricky Math Problem is sweeping the internet with many offering various opinions on how to solve the problem that has social media in a frenzy trying to figure out the solution as only 50% of people will get it right!!!...but can you guess it?.

http://i.dailymail.co.uk/i/newpix/2018/04/25/14/4B8B9CB100000578-5656131-image-a-127_1524664758591.jpg
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I can probably solve it, but I'm so freaking tired right now.

I'd have to reach into my bag of geometry rules, which I haven't used in ages

I'm not sure how you can solve it without hard numbers since you'd only end up with different variables.
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Its clearly 1

aHappySacka posted...

I'm not sure how you can solve it without hard numbers since you'd only end up with different variables.

You can determine the proportion even if you just use say, "x," as a variable.
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aHappySacka posted...

I'm not sure how you can solve it without hard numbers since you'd only end up with different variables.

probably not the best way but if you "slide" the top of the non-diagonal line left and right, all the way to the left, the pink area would be 1/2, and if you slid it all the way to the right, the pink area would be 1/4. so whatever the answer, it must be between 1/2 and 1/4. I don't know for sure that half way between the two would be exactly 1/3, but it's the only answer between 1/2 and 1/4.
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Well, the diagonal line from top left to bottom right cuts the shape in half.
Half of one half would be 1/4, but the shaded area clearly takes up more than half of its side of that diagonal line.
So it's some fraction between 1/4 and 1/2. Too rusty on geometry and too short of cares to try and figure out the answer, but based on the options, it's either 1/3 or 2/5. I'm thinking 2/5...

streamofthesky posted...

Well, the diagonal line from top left to bottom right cuts the shape in half.

See, the thing is that we don't know that for sure since it isn't marked off. You are just making an assumption because it looks like that way.

Sahuagin posted...

aHappySacka posted...

I'm not sure how you can solve it without hard numbers since you'd only end up with different variables.

probably not the best way but if you "slide" the top of the non-diagonal line left and right, all the way to the left, the pink area would be 1/2, and if you slid it all the way to the right, the pink area would be 1/4. so whatever the answer, it must be between 1/2 and 1/4. I don't know for sure that half way between the two would be exactly 1/3, but it's the only answer between 1/2 and 1/4.

Half way between the two would be 3/8

Anyway, couldn't someone measure the damn thing and use actual math. We don't have any actual numbers, but you could damn well get them...
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LinkPizza posted...

Half way between the two would be 3/8

no, I mean I don't know that sliding the line halfway would result in any particular number, 3/8, 1/3, or whatever, I only know that it's between 1/2 and 1/4.
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didn't anyone do their hypotenuse?
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Lokarin posted...

didn't anyone do their hypotenuse?

I bet you're high on potenuse...
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Lokarin posted...

didn't anyone do their hypotenuse?

Bareback, ok?
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LinkPizza posted...

Anyway, couldn't someone measure the damn thing and use actual math. We don't have any actual numbers, but you could damn well get them...

counted wrong at first but...

square is 370x370 pixels so its area is 136,900
height of triangle is about 247?
area of triangle is 247 * 370 / 2 = 45,695
45,695 / 136,900 ~= 33.38%

close enough that whether or not you include a pixel makes a big difference
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Sahuagin posted...

LinkPizza posted...

Anyway, couldn't someone measure the damn thing and use actual math. We don't have any actual numbers, but you could damn well get them...

counted wrong at first but...

square is 370x370 pixels so its area is 136,900
height of triangle is about 247?
area of triangle is 247 * 370 / 2 = 45,695
45,695 / 136,900 ~= 33.38%

close enough that whether or not you include a pixel makes a big difference

That's pretty close to one of the answers...
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My quick answer is 1/3rd.

The lines on the top show the second line is halfway across the top, which should split the 45 degree angle in half, making the two bottom angles 22.5 and 67.5. Net result is that the square is initially cut in half (50% gone), and then of the remaining 50% 67.5% is shaded (or aproximately 33.75% total).

So the answer should be about 1/3rd
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5/7
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I am guessing 1/3.

I want to say where the two lines intersect is roughly 2/3 from the bottom meaning the rectangle that the shaded area is in is 2/3 of what I assume is a square. From there I bisected the shaded area in half forming two rectangles where the shaded area is half of each rectangle meaning the shaded area is half of the 2/3 thus it being 1/3.

Of course I was only assuming where the two lines intersect is 2/3 from the bottom. I'd imagine using the notches that I assume indicate the quarter marks is a more accurate approach even if we assume those notches are quarter marks.
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Stahp with the "approximately", there's a literal solution!
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it's 1/3
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Lokarin posted...

Stahp with the "approximately", there's a literal solution!

Drawing it to scale in AutoCAD? I was tempted to draw this up with the student version that I have installed. I do feel we need more information though. Hell, there's nothing confirming it is a square except for us eyeballing it and are those notches on top actually quarter marks?
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InfestedAdam posted...

are those notches on top actually quarter marks?

Those notches mean they are equal in length...
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Lokarin posted...

InfestedAdam posted...

are those notches on top actually quarter marks?

Those notches mean they are equal in length...

Oh yeaaa.....bloody hell I can't recall when I learned that but I can agree with you. Mighta been in an architect class in high school or geometry. I completely forgot.
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The answer has to be between 1/4 and 1/2, obviously. The answer is also slightly less than 3/8, and 1/3 is the only such answer there, so I'm going with 1/3. Slightly more than 1/4 of the 1/2 triangle isn't shaded--if you were to draw a line from the bottom-left corner to the two opposite sides and the opposite corner, the two end triangles would be a little bigger than the two middle triangles.

1/3 is the only answer choice that it can be, though I'm not entirely sure how to figure that exact fraction out, as opposed to something like 34/97 or something crazy like that.

Also, Ducky, you're terrible at coming up with good fakes. 3/8 would probably be the best-looking wrong answer, yet you didn't include that as a choice for some reason.
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Assuming that all of the 4 sides of the rectangle are the same length:

The smaller triangle on the left has one side of length 1, and another side of length 0.5. Using those you can calculate the three angles of that triangle.

The angle on the top right of that triangle is equivalent to the angle in the bottom left of the shaded triangle.

The angle in the bottom right of the shaded triangle is 45 degrees.

The length of the bottom edge of that triangle is 1.

Using the two angles and the length of that side, you can calculate the third angle and the lengths of the other two sides. Use the length of one of the two sides to calculate the altitude of the triangle, which can be used to calculate the area.

http://www.calculator.net/triangle-calculator.html?vc=&vx=&vy=&va=63.435&vz=1&vb=45&angleunits=d&x=84&y=26

The answer is 1/3
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Algebraic proof:

Consider the shaded triangle to be x, the left triangle y, the top triangle z, and the area of the square 1.

x+y = 1/2, and y+z = 1/4
By rearranging, we get y = 1/2 - x = 1/4 - z
So 1/4 + z = x

x and z are similar triangles, and the base of z is half the length of the base of x, so the area of x is 4 times the area of z, i.e., x = 4z

Combining the two equations, 1/4 + z = x = 4z
1/4 = 3z
z = 1/12

Plug back in to get y = 1/4 - 1/12 = 1/6,
x = 1/2 - y = 1/2 - 1/6 = 1/3
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This looks like something you could solve with calculus as well, purely with numbers on paper and no geometrical knowledge.

Rotate the picture 90 degrees counter-clockwise so that the top side with the notches is now your y-axis. Now you have two functions.

f(x1) = 1*(x1) + 0 = x1
f(x2) = 1/2*(x2) + 1/2 = x2 / 2 + 1/2

Now, you'd have to derive the area highlighted somehow using these lines as functions, but once you did it'd be easy to integrate.
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robrobd posted...

Algebraic proof:

Consider the shaded triangle to be x, the left triangle y, the top triangle z, and the area of the square 1.

x+y = 1/2, and y+z = 1/4
By rearranging, we get y = 1/2 - x = 1/4 - z
So 1/4 + z = x

x and z are similar triangles, and the base of z is half the length of the base of x, so the area of x is 4 times the area of z, i.e., x = 4z

Combining the two equations, 1/4 + z = x = 4z
1/4 = 3z
z = 1/12

Plug back in to get y = 1/4 - 1/12 = 1/6,
x = 1/2 - y = 1/2 - 1/6 = 1/3

I've been out of school too long, but that's a good explanation. I knew 1/3 was the only answer that made sense of the choices given, but I couldn't figure out exactly how to arrive at that specific number.
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Just curious as to why the poll has duplicate answers? 1/2 and 2/4

OS 12TR posted...

Just curious as to why the poll has duplicate answers? 1/2 and 2/4

Because good ol' Ducky is terrible at coming up with good fakes.
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I just picked 1 without looking. Im really, astonishingly bad with this sort of thing. after looking at it, I'd change my answer to "Yeah no idea."
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Who picked 1? Unless they thought it meant how many parts are shaded, I'm not sure how they arrived there...
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1/4
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When it comes to think kind of question, my mind is blur. I need someone to teach me math again.
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I treated the four corners as coordinates on a grid and then described the two crossing lines as functions.

The line cutting the square in half would be y = 1 - x.
The second line would be y = 2x.

By taking 1 - x = 2x and solving for x I got the x coordinate for the intersection, 1/3.
Put 1/3 into one of the functions to get the y coordinate which turns out to be 2/3.

Draw a line straight down from the intersection and the result is two triangles.
One with the sides 1/3 & 2/3, and one where both sides are 2/3.

The sum of their areas is (1/3 * 2/3)/2 + (2/3 * 2/3)/2 which simplifies to 1/9 + 2/9 which can again be simplified into 1/3.

Edit: It's really fun to see how people solve it using different methods.
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Gonna go with 2/5. 1/3 is clearly taken by everyone else.
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K3nnyan posted...

I treated the four corners as coordinates on a grid and then described the two crossing lines as functions.

The line cutting the square in half would be y = 1 - x.
The second line would be y = 2x.

By taking 1 - x = 2x and solving for x I got the x coordinate for the intersection, 1/3.
Put 1/3 into one of the functions to get the y coordinate which turns out to be 2/3.

Draw a line straight down from the intersection and the result is two triangles.
One with the sides 1/3 & 2/3, and one where both sides are 2/3.

The sum of their areas is (1/3 * 2/3)/2 + (2/3 * 2/3)/2 which simplifies to 1/9 + 2/9 which can again be simplified into 1/3.

Edit: It's really fun to see how people solve it using different methods.

Oh thats quite an elegant and straightforward solution, though I wouldnt have thought to approach it that way. I like that.
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And now that I'm more awake and I'm looking at it again I see that I could have just treated the last step as one triangle with the base being 1 and the height being 2/3.. So that would have been more elegant.. Still, it wasn't wrong..
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I know this topic is dead but I was thinking about it last night so hard while I was trying to go to sleep that I was imagining calculus in my head... Anyway, here is the calculus method:
(forget what I said earlier about considering the plot turned 90 degrees)

Let's call the upward line f'(xA) and the downward line f'(xB). The solution is area C. The point where they meet is coordinate (x1,y1). Assume the square is a unit size, 1 x 1. What we want to find is the integral (area underneath) the first line up to the point where they meet (x1), and from that point the integral of the second line until it meets the x-axis at x=1.

C = f(xA) from 0 to x1 + f(xB) from x1 to 1.

f'(xA) = (1/(1/2))x + 0 = 2x
f'(xB) = -(1/1)x + 1 = 1 - x

f(xA) = integrate f'(xA) = x^2
f(xB) = integrate f'(xB) = x - (x^2)/2

Find x1: This is the point where f'(xA) = f'(xB).
2*x1 = 1 - x1
2*x1 + x1 = 1
x1 = 1/3

Solve for C = [f(xA) 0_x1] + [f(xB) x1_1].
C = [(1/3)^2 - (0)^2] + [(1 - (1^2)/2) - (1/3 - (1/3)^2/2)]
C = [1/9] + [1/2 - 5/18]
C = [2/18] + [4/18]
C = 1/3

This is the most similar to K3nnyan's geometric solution which was to solve for the point where they meet and find the area of the two triangles, it's different math but the same concept. To me, robrobd's algebraic solution was the most elegant. My solution here is the most rigorous proof, it doesn't require any geometric knowledge (like how to split a rectangular area into two triangles) or trigonometry like Magus's calculator method. The reason mine is more robust is because the other methods would fall apart if you replaced one of the lines with, say, a circular arc, or some other arbitrary curve.

Edit: This was fun :)
I don't mean to say my method was "better" than anyone and I'm not hating on anyone's math. I agree it's really fun finding all the different ways it can be solved.
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Isn't the answer one fourth because it is a trick question? Only one out of the four fractions is shaded.
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