6 of 100: Sneaky Squares

Here is the intuition: This illustration suggests that the answer is $\frac{1}{5},$ but the illustration alone is not a proof. In order to be sure that those five apparently square shapes are all truly congruent squares, we need to do some extra, careful reasoning.

PROOF:

Part 1: All of the yellow angles are right angles and the 4 big triangles colored above are congruent. Firstly, we are given that the large shape is a square. Therefore the sum of 1 small (dark blue) angle and 1 large (light blue) angle that comprise each corner must be $90^{\circ}.$ Therefore, the four large triangles surrounding the green region are right triangles (the angle closest to the center of the figure is the right angle) and these four right triangles are congruent by Angle-Side-Angle. Finally, we can can conclude that all of the yellow angles in the central region are right angles as they are either the right angle of one of the triangles, opposite to one of these angles, or supplementary to one of these angles.

Part 2: The small triangle pieces that move and spin in the image above are all similar to the large triangles (1:2).

By Angle-Angle-Angle, we can conclude that the four small triangles (the tips of the larger triangles) are similar to the larger triangles and that the scale factor is 1:2. Therefore, the three line segments above indicated with red hash-marks are all equal in length and the segment with the purple hash-mark is half of that length.

Part 3: The 5 shapes made in the image at the top of this solution are all congruent squares.

Last step: spin the small triangles around to make four squares surrounding a fifth central square. We know they're squares because we've shown that all four side lengths of each square are congruent line segments and all four angles of each square are right angles.

This is perhaps not the fastest solution, but it's still one of my favorite ways to visualize and prove that the answer is $\frac{1}{5}.$

$A$ is a midpoint, so $\tan\theta=\dfrac12, \sec\theta=\sqrt{1+\tan^2\theta}=\dfrac{\sqrt{5}}{2}$

Assuming the large square to have a side (and area) 1: $\overline{BD}=\cos\theta=\dfrac{2}{\sqrt{5}}$

$\triangle EDB$ and $\triangle FDC$ are similar and $F$ is a midpoint of $\overline{ED}$ therefore $\overline{BC}=\dfrac12\overline{BD}=\dfrac{1}{\sqrt{5}}$

The area of the square is $\overline{BC}^2=\left(\dfrac{1}{\sqrt{5}}\right)^2=\boxed{\dfrac15}$

By moving the 4 little right triangles to create squares, we can clearly see that the green square is just one of 5 identical squares, and has $\frac{1}{5}$ of the total area.

This is probably an overkill for a "2 thorn" problem, I will attempt a general solution as follows:

From the diagram k will be 0.5 for this question.

The diagonal lines will have length $\sqrt{1+k^2}$ .

Due to the similarity of the triangles ABC and BDE $\frac{x}{1-k} = \frac{1} {\sqrt{1+k^2}}$

Therefore: x = $\frac{1-k}{\sqrt{1+k^2}}$

$x^2 = \frac {(1-k)^2}{1+k^2}$ which is the area of the internal square.

When k=0.5 then the area is $\frac{\frac{1}{4}}{\frac{5}{4}}$ = $\boxed{\frac{1}{5}}$

Other values:

Total area of our square is $2 \times 2 = 4$

On each of the four sides, we have a $1 \times 2$ right triangle, with a hypotenuse of $\sqrt{5}$ .

Taking out two of those triangles leaves us with a parallelogram, with one side equal to 1, the other to $\sqrt{5}$ . To find it's area, we can use 1 as the base, and the width of the square, 2, as height. So the area of the parallelogram is 2 (which makes sense, as we've already removed two triangles each with an area of 1 from our square).

We can say that a side of the green square equals the height of the parallelogram where the base is $\sqrt{5}$ . So $\sqrt{5} \times h = 2$ .

$h = \frac{2}{\sqrt{5}}$

The area of the green square is $\frac{4}{5}$ , which is $\frac{1}{5}$ of the larger square.

Counting them all, just for kicks!

Using coordinate geometry: Let scale be such that large square has sides of length 4. Equations of diagonal lines: $y = 4 - 2x, y = 4 - 2(x-2), y = (1/2) x, y = (1/2)x + 2$ . These equations define the four points of intersection that form the for corners of the smaller square. Simple algebra shows that, for example, the lowest corner is $(8/5,4/5)$ and the leftmost corner is $(4/5,12/5)$ . Use the Pythagorean theorem to find the area of the central square: $(4/5)^2 + (8/5)^2 = 16/5$ . The area of the large square is $16$ , so the smaller square has $1/5$ the area of the larger square.

How about a simpler solution?

We know that the edge of the green square is less than the length of half the larger square's edge. How?

Pythagoras theorem.

Therefore, it must be less than $\frac{1}{4}$ , and with the given answer choices - it must be $\frac{1}{5}$ .

The End.

Sorry its bit small I'm new here. Control + Scrollwheel up to magnify your screen may help. All trigonometry! Bonus question answered trig and series. Additional bonus solution:

The $IJKL$ is a square (because symmetry), so $AIE\angle =BJF\angle=AJB\angle =ABF\angle =90°$ . If $BAF\angle =\alpha$ and $90°-\alpha=\beta$ , then $AFB\angle =\beta=AEI\angle =ABJ\angle$ and $JBF\angle =\alpha$ . The $AEI,\space ABJ, \space ABF$ right-angled triangles are similar. Let the length of the square's side is $2$ (it doesn't matter). The triangles' areas' proportion is equal to the hypotenuses' proportion. So $T_{AEI}:T_{ABJ}:T_{ABF}=AE^2:AB^2:AF^2=1:4:5$ (because using the Pythagoras theorem we get $AF=\sqrt{AB^2+BF^2}=\sqrt{5}$ ). We can see that $T_{ABF}=AB*BF/2=1$ , so by calculating we get that $T_{ABJ}=0.8$ . So $T_{IJKL}=T_{ABCD}-T_{ABJ}-T_{BKC}-T_{CLD}-T_{DIA}=4-4*T_{ABJ}=0.8=\boxed{\frac{1}{5}}$ (because symmetry).

Just visually, you can tell it must be less than 1/4 and there's only one available answer that is less than 1/4.

My diagram illustrates similarity. We wish to calculate the length of the shorter blue line, so we can set up a proportion. Let the length of the longer blue line be $1$ . Then the length of the longer red line is $\dfrac{1}{2}$ and by the Pythagorean theorem, the length of the longer green line is $\dfrac{\sqrt{5}}{2}$ . The length of the shorter green line is $\dfrac{1}{2}$ . We set up a proportion using long blue line: long green line equals short blue line:short green line. Finally, we want the side length of the inner square. We know that this is the same as the length of the short blue line, because if you look at the right triangle whose hypotenuse is the long blue line, we see that the long blue line is twice as long as the hypotenuse of the right triangle whose hypotenuse is the short green line, and we also can deduce that these triangles are similar by using the AA similarity theorem. Thus, any leg in the smaller triangle has half the length of its corresponding leg in the larger triangle. All of this implies that the side length of the smaller square is $\dfrac{\sqrt{5}}{5}$ , yielding an area of $\boxed{ \dfrac{1}{5} }$ of the total area of $1$ .

A big square had 5 pieces of small square. So the fraction of the green square is 1/5

Making the big square have length 2 and the bottom left corner be the origin of the x,y coordinate system, we have:

• The line with positive slope that crosses the origin has equation: $y=\frac{1}{2}x$ ;
• The negative slope line that crosses $(0,2)$ has equation: $y=2(1-x)$ ;
• The other negative slope line has equation: $2(1-x)+2$ .

Solving to see when the negative slope lines intersect the positive slope line, we find the coordinates of the two bottom cornes of the small triangle, they are: $\left(\frac{4}{5},\frac{2}{5}\right)$ and $\left(\frac{8}{5},\frac{4}{5}\right)$ . Using Pythagoras we get that the area of the small square is $\frac{4}{5}$ , that is $\frac{1}{5}$ of the area of the great square.

...

I wish I had the geometric idea instead.

It is evident that the side length of the green square is less than half of the side length of the larger square. The area of the green square is therefore less than the area of one quarter of the larger square. The only answer choice less than $\frac{1}{4}$ is $\frac{1}{5}$ .

Because I was lazy, I tried to solve this without any trig or geometry, just by looking at the list of answers. It is obvious that the length of the side of the green square is less than 1/2 of the length of the side of the large square (which is the angled side of the trapezoid). Therefore the green square must be < 1/4 of the size of the large square. The only answer that is < 1/4 is 1/5.

The general formula would be: $\quad f(n)=\frac { 1 }{ { 2n }^{ 2 }-2n+1 }$ where n in the denominator and when numerator is 1. (Meaning joins to $\frac { 1 }{ n }$ )

Proof: We are trying to find x

We know that:

$cos(\theta )=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+({ \frac { n-1 }{ n } ) }^{ 2 } } } =\frac { n }{ \sqrt { 2{ n }^{ 2 }-2n+1 } } \\(Above\quad equation \quad is\quad from\quad the\quad big\quad triangle \quad at \quad the\quad bottom) \\ x=\frac { 1 }{ n } cos(\theta )=\frac { 1 }{ n } .\frac { n }{ \sqrt { 2{ n }^{ 2 }-2n+1 } } =\frac { 1 }{ \sqrt { 2{ n }^{ 2 }-2n+1 } }$

The area of the square would be:

$Area = { x }^{ 2 }= \frac { 1 }{ 2{ n }^{ 2 }-2n+1 }$

Now, to answer the question, just sub in x=2 into the formula.

Hopefully this makes sense. If I have made a mistake, please let me know.

In the above problem, there are 4 big right angled triangles with heights $y$ , with base equal to the height of the smaller triangle, and with hypotenuse 1 (Take the side of the square to be 1; we're anyways finding out what fraction of the area the square occupies), and the smaller right triangles with heights $x$ and bases that we really don't care about. So, by AA similarity, $\dfrac{x}{y} = \dfrac{\frac{1}{2}}{1} \implies y = 2x$ .

By Pythagoras, $x = \dfrac{1}{\sqrt{5}}$ . Observe that $x$ is also the side of the square. Hence the area of the square is $A = \dfrac{1}{5}$ . Which is the required result.

Here is the general solution. See figure below for definitions of symbols:

$R = \sqrt{1+y^2}$

By the law of similar triangles we have:

${\displaystyle x\over \displaystyle y} = {\displaystyle 1\over \displaystyle R}$

so

$x = {\displaystyle y \over \displaystyle \sqrt{1+y^2}}$

$A = x^2 = {\displaystyle y^2 \over \displaystyle 1 + y^2}$

If we set $y = 1/n$ , then we have:

$A(n) = {\displaystyle 1 \over \displaystyle 1 + n^2}.$

So $A(2) = 1/5, A(3) = 1/10, A(4) = 1/17, A(5) = 1/26.$

Let's consider a 1x1 square.

Because of Thales' theorem (or Intercept theorem), $AL = LK = x$ , so we can write that (according to an Euclid's theorem, in Italy it is called "First Euclid's Theorem"):

$AB^2 = AK \cdot AF \rightarrow 1^2=2x \cdot \sqrt{1^2+(\frac{1}{2})^2} \rightarrow 1=2x \cdot \frac{\sqrt{5}}{2}$

$x=L_{square}=\frac{1}{\sqrt{5}}$

So the area of the square will be $L_{square}^2=\frac{1}{5}$

I put it on the coordinate plane and made equations of 3 of the lines. Then I found the intersection points of the lines so that I had the coordinates of 2 consecutive corners of the square. Once I had that I used the distance formula to find the length of the side of the green square, then divided the area of the green square by the area of the whole square.

I did it in a different way, You can take any arbitrary positive integer as side of large square. I took the large square side as 10 units. then the midpoint divided sides into 5 units. There are many right triangles. Taking the length of arm 4 and 3, it satisfies the Pythagoras theorem. After finding length of each segment, find area of each part, subtract from total area which will give you the ratio (approx) = $\dfrac{1}{5}$