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This brain teaser turned out to be a brain boggler. As I am the type of math need that dwells on a single problem until it gas been solved ( and understood ).

I don't think there is enough information given to find the area of shaded region . I tried drawing lines to make congruent pieces of the shaded region in terms of a side of the square call it $x$. I was even attacking it with trig to see if angles given since there are many parallel sides, but still no avail. Please help.

enter image description here

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  • $\begingroup$ May I ask where this nice problem was found? $\endgroup$
    – mickep
    Mar 3, 2018 at 18:29
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    $\begingroup$ mickep maybe u would be so inclined to upvote my question ? and I believe it was on pint5$erest . it was wierd that no solution was on that site either , and no where to be found anywhere on the net ( I searched many different wordings on Google of the problem and found the empty set 😃 $\endgroup$
    – Randin
    Mar 4, 2018 at 7:14

2 Answers 2

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Consider the following illustration:

enter image description here

Write top angle (smaller one) as $\theta$ and side of square as $s$. Then use the fact that projection of red and green lines add up to side lengths of rectangle.

In the image, green line is for first equation and red line for second.

$$(3s+s \tan(\theta)) \cos(\theta) = 11 \tag{1}$$

$$(3s+2s \tan(\theta)) \cos(\theta) = 13 \tag{2}$$

From here we get $s \sin(\theta) = 2$ and $s \cos(\theta) =3 $. So $\tan(\theta) = \frac{2}{3}$. Then $s = \sqrt{13}$.

The required area is $13 \cdot 11 - 6s^2 = 13\cdot 5 = 65$

Note: The redline does not touch the upper side at corner. It is just meant to show that we have to take projection of this line along vertical side of length 13. Just to show that red line makes angle $\theta$ with vertical side.

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  • $\begingroup$ looks sound. but i was wondering if it can be done without any trig as I geometry as well .that would be the best trig approach I'm thinking $\endgroup$
    – Randin
    Mar 2, 2018 at 14:58
  • $\begingroup$ Hmm, sounds interesting, I would also wish to see. This one came naturally though. $\endgroup$
    – King Tut
    Mar 2, 2018 at 14:59
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    $\begingroup$ Denote the longer projection of a white square side by $a$, and the shorter by $b$. Then $2a + b + a + b = 13$ (left side of the big square) and $2b + 2a - b + a = 11$ (upper side), from which $b = 2$ and $a = 3$. Now, by Pythagoras' theorem, the area of a white square is $a^2 + b^2 = 9 + 4 = 13$, hence the white area is $78$ and the shaded is $65$. $\endgroup$
    – Kolya Ivankov
    Mar 2, 2018 at 16:04
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    $\begingroup$ @King Tut Of course it's based on your solution, just making it plausible for a Pythagoras :) $\endgroup$
    – Kolya Ivankov
    Mar 2, 2018 at 16:39
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    $\begingroup$ Nice. It means that the shaded area is as large as 5 white squares. I wonder if it's possible to cut this area in small pieces and rearrange them to get 5 squares. $\endgroup$
    – Eric Duminil
    Mar 2, 2018 at 22:22
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By a recommendation of @King Tut, adding my comment as an answer:

Denote the longer projection of a white square side by $a$, and the shorter by $b$. Then $2a+b+a+b=13$ (left side of the big square) and $2b+2a−b+a=11$ (upper side). From that we get $b=2$ and $a=3$. Now, by Pythagoras' theorem, the area of a white square is $a^2+b^2=9+4=13$, hence the white area is $78$ and the shaded is $65$.

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