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I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:

What is the blue area?

Brilliant Image

Hint: Think outside the box

My answer:

I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:

My attempt

This then gave me the following equation to solve:

\begin{align}y^2&=2+3+4+x\\ &\Downarrow\\ x&=y^2-9\end{align}

And the following equations to do so: \begin{align}\frac {ya}2 &=4\\ \frac {bc}2 &=3\\ \frac {yd}2 &=2\\ a+b&=y\\ c+d&=y\end{align}

I solved these to obtain:

$$a=2, b=2, c=3, d=1, y=4$$

And thus $$x=4^2-9=7$$

My question:

Is there another way I could have solved this, using the hint to think outside the box?

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12 Answers 12

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Not a completely geometric solution and I am assuming that only integers are allowed.

Since area of yellow triangle is twice the red one, so $a=2d$. Now divide the square into four rectangles by drawing horizontal line from point between $a$ and $b$ and a vertical line from point between $c$ and $d$. Then $$4+8+6-a.\frac{a}{2}=y^2$$ which gives $$a^2+2y^2=36.$$ This has solution $a=2, y=4$.

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  • $\begingroup$ This is a nice solution, although I'm still not sure it completely relates to the hint $\endgroup$ – lioness99a Jun 15 '17 at 13:30
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The hint may assume two things: to draw extra lines 1) outside; 2) inside the box so that the solution is simple to understand.

Here is yet another method: draw the lines to divide the square into four rectangles indicated by $A, B, C$ and $3+3$:

enter image description here

$$\begin{cases} A+B=8 \\ A+C=4 \\ CB=6A \end{cases} \stackrel{(1)-(2)}\Rightarrow \begin{cases} B=C+4 \\ A=4-C \\ C(C+4)=6(4-C) \end{cases} \Rightarrow $$ $$C^2+10C-24=0 \Rightarrow C=2, A=2, B=6$$

Hence: Blue area = $(A+B+C+6)-(2+3+4)=16-9=7.$

P.S. I wonder what brilliant method the proposer (on the brilliant.org) has.

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  • $\begingroup$ Agreed, I haven't been able to find the link again as I saw it in an advert $\endgroup$ – lioness99a Jun 17 '17 at 17:14
  • $\begingroup$ Don't worry, the link doesn't go to a page which answers the question. $\endgroup$ – Trejkaz Jul 14 '17 at 11:23
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Here is an alternative solution. See the graph for labeling: enter image description here

From the trapezium area formula we get: $$\frac{a+x}{2}\cdot x=5+A \Rightarrow ax=10+2A-(9+A) \Rightarrow ax=A+1 \ \ (1)$$ $$\frac{b+x}{2}\cdot x=7+A \Rightarrow bx=14+2A-(9+A) \Rightarrow bx=A+5 \ \ (2)$$ $$ab=6 \ \ (3)$$

Now multiply $(1)$ and $(2)$ and substitute $(3)$: $$abx^2=(A+1)(A+5) \Rightarrow 6(A+9)=(A+1)(A+5) \Rightarrow A^2=49 \Rightarrow A=7.$$

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  • $\begingroup$ Thanks, but how does this use the hint? $\endgroup$ – lioness99a Jun 15 '17 at 12:01
  • $\begingroup$ Still very algebraic. There must be some kind of brilliant geometric insight eluding us all. $\endgroup$ – Deepak Jun 15 '17 at 12:02
  • $\begingroup$ But it does avoid a biquadratic solution, so it's better than mine, so I'm giving you an upvote. $\endgroup$ – Deepak Jun 15 '17 at 12:17
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We may assume that the bounding rectangle is a square with side $s$. Guessing $s=4$ lets everything fall into place, and the resulting area of the blue triangle is $7$.

Now it is easily seen that making $s<4$ and keeping the areas $2$ and $4$ for the red and the yellow triangles would decrease the area of the green triangle. Similarly, making $s>4$ would increase the area of the green triangle.

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Until someone posts with "brilliant geometric insight" I was hoping exists, I might as well post my solution. Essentially similar to the OP's, except I used fewer variables.

I assigned $x$ to be the base of the area $2$ triangle. That meant the height of the same triangle would be $\frac 4x$. The dimensions of the $4$ triangle would be $\frac 4x$ and $2x$. And that of the $3$ triangle would be $\frac 4x - 2x$ and (after some simplification $\frac{3x}{2-x^2}$.

Based on the lower (southern) border of the square, setting up the equation $\frac 4x = x + \frac{3x}{2-x^2}$, we get a biquadratic $(x^2 - 8)(x^2-1) = 0$. $x = 2\sqrt 2$ can be dismissed as a solution because it makes dimensions of the $3$ triangle negative, leaving only $x = 1$. Very quickly, we can figure out that the area of the square is $4^2 = 16$, and therefore the blue area is $16 - 2-4-3 = 7$.

I'm really hoping there's a simpler, far more elegant solution.

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  • $\begingroup$ That's definitely a slightly nicer method than mine, but I agree, there has to be some sort of geometric insight we're all missing for now $\endgroup$ – lioness99a Jun 15 '17 at 12:25
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if the square side is x, then the width of area 2 is 4/x and the reminder of that side is (x^2-4)/x. Similiarly the width of the 4 area is 8/x, and the other side of the 3 area is (x^2-8)/x. Then from the 3 area (x^2-4)/x * (x^2-8)/x = 3*2. Let A = x^2, then (A-4)(A-8) = 6A. A^2 -18A +32 = 0 ... A = 16, and blue = 7

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I honestly just eye-balled this one, and came up with a solution without verifying it was the only solution:

Since the outside edge shared by Green and Yellow appeared roughly at halfway point, and the side shared by red and yellow at the $\frac34$ point, it seemed obvious to me that each of the full square edges likely had a length of $4$, the green-yellow side was split into $2$ each, and the red-yellow was split $1-3$. That satisfied Greens $\frac{4 \times 2}2$, Yellows $\frac{3\times 2}2$, and Reds $\frac{4\times 1}2$.

And then $$4 \times 4 - (4 + 3 + 2) = 7.$$

But I still would have liked the clear statement that the puzzle itself was a proper square, and not some other quadrilateral. I am satisfied with all y'alls math that this is the only solution, but only with the starting assumption that the figure is a proper square.

Would it have killed you to add a couple right angle symbols, and the hash marks showing that the outside full length edges of red and yellow were equal?

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  • $\begingroup$ Can anyone prove that the enclosing quadrilateral is in fact a square, given only the information asked by the original question? All the solutions here appear to start from that assumption. $\endgroup$ – Brian Gix Sep 21 '17 at 18:39
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You see that hint & wonder if they mean "look outside the box" quite literally. Also, there is no proof that it's a square. But they are about making things easier for people that struggle with maths, not those who can already do it.

This is clearly not meant to be done via lots of tedious algebra. They never do that. I eventually eliminated variables assuming a square to get (a - 1)(a - 2)(a - 4) = 0, & trivially only a = 2 is possible. Therefore trivially b = 2, c = 3, d = 1 & this means it's a square of side length 4.

Once you've found a solution, the question itself implies there is only one solution (find the blue area, not a list of solutions that the blue area can be). So we can get there one way or another in the end.

I think the "brilliant" thing is far simpler than what you're all looking for. There is no quick & easy way. I had a look at some of their other problems and they are usually WAY easier than they seem to be on the surface. Painfully easy most of the time. In other words they are saying "look, it's a square dummy" and yes, they are assuming that you will assume integer solutions. Therefore 16 - 9 = 7. That's all.

Basically, a poor question.

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There is a wonderful generalization of the solution for this problem:

$$A = \sqrt{(a + b + c)^2 – 4ac}.$$

Where $a$, $b$ and $c$ are the known areas, and $A$ is the unknown area.

Caveat: $a$ and $c$ must be the triangles whose long legs are the side of the square, in this case $7$ and $9$.

$$A = \sqrt{729 – 4 \times 9 \times 7} = \sqrt{729- 252} = 21.84.$$

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Green triangle has 2x area of red triangle. Heights of both triangles are equal because all sides of square are equal. so base of green triangle = 2x base of red triangle.So r=length of base of red triangle and 2r=base of green triangle. Let s= length of side of square also height of both triangles. for red triangle 2=1/2rs so rs=4. same relation hold for green triangle. 4=1/2(2rs) so rs=4 again. Referring to the yellow triangle bottom side=s-r and right side=s-2r. So the area of the triangle is 3= 1/2 (s-r)(s-2r) then 6=( s-r)( s-2r) so 6=s^2-3sr+2r^2. since rs=4. r=4/s. substitute in equation. then 6=S^2-3s(4/s)+2(16/s^2) 6=s^2-12+32/s^2 , 18=s^2+32/s^2, 18s^2=s^4+32, s^4-18s^2+32=0, Let x=s^2, then x^2-18x+32=0, Factor (x-16) (x-2)=0 x=16, x=2 disregard x=2 x=16=s^2 then 16-4-2-3=7 area of Blue triangle

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Assume that the hypotenuses of the yellow and green triangles bisect the right side of the quadrilateral, which is assumed to be a square with side s. Area formula is (1/2)bh where b=s and h=s/2, so the area of the yellow triangle is equal to ss/4, which is equal to 4. ss is thus equal to 4*4=16, so s=4. The area of the square is also equal to s*s, which is 16. The blue triangle is thus equal to 16-4-3-2 = 7.

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    $\begingroup$ Assume that the hypotenuses of the yellow and green triangles bisect the right side of the quadrilateral Why assume that? $\endgroup$ – dxiv Jul 14 '17 at 3:59
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From the diagram below, you can see that triangle GHK = 3. You also see that triangle FGK = 0.5. Finally, you can see that triangle AHE = 4, but we have to subtract out triangle AFE (since it is red). Triangle AFE = 0.5. That means the blue area is 3 (GHK) + 0.5 (FGK) + 4 (AHE) - 0.5 (AFE) = 7. I believe this is "outside the box" as no "equations" are used, only simple addition and subtraction.

redrawn box

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    $\begingroup$ triangle FGK = 0.5 How do you "see" that? It's not given that $H$ is the midpoint of $LD\,$. $\endgroup$ – dxiv Jul 14 '17 at 3:53
  • $\begingroup$ Not assuming that H is the midpoint of LD. I am assuming that ADLI is a square, though. Since both triangles red and green have one side of the square as a base and the area of green is twice that of red, DH is twice IK. IK is 1/4 of the base (AI, side of the square). Since DH is twice IK, it is 2 x 1/4, or half of the side of the square. Therefore, since DH is half of DL, so DH = HL, which means H is the midpoint of DL. $\endgroup$ – Biro Jul 14 '17 at 6:32
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    $\begingroup$ IK is 1/4 of the base Where does that follow from? $\endgroup$ – dxiv Jul 14 '17 at 6:35

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