# Area of a square inside a square created by connecting point-opposite midpoint

Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each.

$H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively.

What will the area of the square formed in the middle be?

I know that this problem can be solved by trigonometry by using Area of triangle ($\frac{1}{2}ab\sin{c}$) but, is there another method or visual proof?

• Hint. Rotate triangle with edge $DF$ clockwise half a turn. Nov 13, 2017 at 14:51
• this is an old math olympiad problem Nov 13, 2017 at 14:57
• @Arthur Sorry i am editing it now Nov 13, 2017 at 14:58
• See also: Finding the area of shaded region Nov 15, 2017 at 6:02
• @EthanBolker: Around $F$. ;) Nov 15, 2017 at 11:15

By moving small triangles we can make $5$ equal small squares.

• How do you know that moving 1 to 2, you obtain a square (or, in other words, that the segments connecting the vertices of the squares have the same length as the inner square side)? Nov 13, 2017 at 20:40
• @OxTaz, those triangles 1 and 2 are right triangles, their hypotenuse are equal and their opposite angles are congruent. Therefore triangles 1 and 2 are equal. Nov 13, 2017 at 20:59
• How do you know that long edge from corner of inner square to corner of outer square is equal to length of side of blue square? That does not seem obvious to me. Nov 13, 2017 at 22:06
• @Jiminion, Lines are drawn from vertices of the main square to the middle of the opposite sides. In the large right angled triangle (triangle 1 and a part of small square) the line is parallel to the small leg divides both the hypotenuse and the other leg into two equal parts. Nov 13, 2017 at 22:23
• @Jiminion I had the same issue. Then I noticed that the long leg of triangle 1 starts as the "height" of a new square, but after the rotation it is the "width" of a new square, thus it is in fact a square. (And after rotation the "width" of the new square is clearly equal to the side-length of the blue square). Nov 15, 2017 at 19:42

By drawing lines along and parallel to the segments within the square, a grid overlapping the original square is produced. If $A$ represents the area of the shaded square, then $9A$ represents the area of the circumscribing square with sides parallel to the shaded square

It is easy to see a right triangle on each side of the original square with its hypotenuse on the side of the original square, each with an area of $\frac{2\cdot A}{2} = A$. Thus the total area of the original square is $5A$, or the ratio of the area of the shaded square to the original square is $1/5$.

Let $GK$ be a perpendicular to $AF$, $GK=x$ be the side-length of the little square

and $BH\cap AF=\{M\}$.

Thus, $AM=GK$ and since $\Delta AMH\sim\Delta ADF$, we obtain: $$\frac{x}{1}=\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}},$$ which gives $x=\frac{1}{\sqrt5}$ and the answer: $\frac{1}{5}$.

• An interesting appendix to this answer is that if $AH=DF=CE=BG=\frac{a}{b}$ then the area turns out to be $$\frac{(a-b)^2}{a^2+b^2}.$$ Nov 14, 2017 at 14:44
• Yes! It's beautiful, John! Nov 14, 2017 at 15:54
• I don't get the "thus AM = GK". I had the same problem with seyed solution. Are you saying AKG triangle is same as AMH? Nov 14, 2017 at 19:30
• @Jiminion Because $\Delta AHM\cong\Delta GAK$.By two angles and the side. Nov 14, 2017 at 19:41
• @MichaelRozenberg OK, I can see that. Nov 14, 2017 at 20:05

Triangle BCG is $\frac 12-1-\frac 12\sqrt 5$ with area $\frac 14$.
The small right triangles are similar with hypotenuse $\frac 12$ so area $\frac 14\frac {(\frac 12)^2}{(\frac 12\sqrt 5)^2}=\frac 1{20}$.
To get the area of the center square you take the square, subtract four triangles like BCG, and add four small ones that you subtracted twice, to get $1-4\cdot \frac 14+4\cdot \frac 1{20}=\frac 15$

• I was using the same method! Although the other answer answer is also good (it uses similar similarity for finding length of sides but is more direct.) Nov 13, 2017 at 15:03
• This is the only one that doesn't make (unproved) assumptions, in my opinion. Nov 14, 2017 at 16:32

Here's a solution using analytic geometry which doesn't require any particular insight:

Set up a coordinate system such that $B = (0, 0)$ and $C = (1, 0)$. Then the equation of line $BH$ is $y - 2x = 0$, and the equation of line $CG$ is $2y + x = 1$. Therefore, their intersection is $(\frac{1}{5}, \frac{2}{5})$. Similarly, the equation of line $ED$ is $y - 2x = -1$, so the intersection of lines $ED$ and $CG$ is $(\frac{3}{5}, \frac{1}{5})$. This gives that the side of the inner square is the distance between these two points, which is $$\sqrt{\left(\frac{1}{5} - \frac{3}{5}\right)^2 + \left(\frac{2}{5} - \frac{1}{5}\right)^2} = \frac{1}{\sqrt{5}}.$$ Therefore, the area of the square is the square of this length, or $\frac{1}{5}$.

Why don't you move the four small triangles in such a way to construct a "cross" made of five equal squares?

In $\triangle ABC$ and $\triangle DCF$

$AC=CF$

$DF=BC$(Pythagoras)

$\angle BAC=\angle DCF$ , by $RHS$ congruency they are congruent. And similarly, all such triangles are congruent $\triangle ABC$= $\triangle CDF$=$\triangle FKL$=$\triangle LZA$

Consider $\angle ECD=x$ , $\angle ECK=90-x$ and $\angle JFK=x$ (Why? CPCT)

Therefore, $\angle FEC=90^\circ$. Similar argument for $\angle FJK=90^\circ$

$EC || JK \implies BC||LK$ and we can argue the same for other two lines, too .

And we get $AZ||DF$.

Consider $\triangle DEC$ and $\triangle LYZ$

$\angle LYZ=\angle DEC=90^\circ$

$\angle YZL=\angle EDC$ (Since they are parallel)

$DC=LZ$

Therefore, the two triangles are congruent.

Now consider $\triangle LYZ$ and $\triangle ABX$

$AB=LZ$

$\angle AXB=\angle LYZ=90^\circ$

$\angle BAX=\angle YLZ$ (Why?)

Therefore the triangles are congruent.

This proves that all the four smaller triangles are congruent.

Congruency for trapezoids,

$LY+YJ+JK=BX+XE+EC \implies XE=YJ$

$YZ=DE$, $AD=ZF$ and $AX=JF$ (From congruency of triangles)

And $\angle FJY=\angle AXE=90^\circ$

By ASSSS congruency, they are congruent. Similarly, all four are congruent.

$XY=EJ=XE=YJ$ (Since trapezoids are congruent), Therefore that's a square (All are right trapezoids)

$YZ=JK$ and $YZ+JK=JF$. When you rearrange the shapes (Trapezoid+triangle), you get four congruent squares. And these squares are congruent to the square $XYJE$.

Therefore area$(XYJE)= \dfrac{1}{5} \times (ACFL)=\dfrac{1}{5}$ square units

• This is nice, but where do you deduce from that the four trapezoids are congruent? Mar 26, 2013 at 14:45
• @DonAntonio: Consider these points: $LY+YJ+JK=BX+XE+EC \implies XE=YJ$. And $DE=YZ=AX=JF$ and $\angle AXE= \angle YJK$. From ASSSS congruency, they are congruent. Similarly, you can deduce the other two. Right? Mar 26, 2013 at 14:53
• @DonAntonio: You can argue the small triangles are congruent, then all the sides of the trapezoids are equal and you have the two right angles. Mar 26, 2013 at 14:55
• Oh, I know that, Ross, but the OP seems to be struggling with this question, so only state that this or that is congruent/similar may not help him/her much. Mar 26, 2013 at 16:29
• Inceptio, that's fine. My whole point in my comments to all was in making it sure that the OP understand that there's something that needs to be proved and that what was being argued isn't obvious or immediate. Way to go for your effort and, of course, +1 Mar 26, 2013 at 18:03

Since it is a square,each (triangle + trapezium = square) . Hence the bigger square is made up of 5 smaller squares. Hence area = 1/5.

• What "triangle" plus what "trapezium" equals what "square"?...and why, of course? Mar 26, 2013 at 14:30
• I think you did not see the term 'each'. In English each means all the things. Since there are 4 triangles and 4 trapeziums, the rest is implied.
– lsp
Mar 26, 2013 at 14:34
• There are lots of triangles in that figure, and at least two squares, so my question remains (the four trapeziums are congruent, though) Mar 26, 2013 at 14:36
• @DonAntonio Think of A, C, L and F each having a 'hinge' that can flop the triangle at each hinge clockwise onto the trapezium. In the case of the BADEX structure, AB = AD (new) angle XAB (after folding) will be 90 and sides AX = AX' where X' is the new point created by the folding. So you have 2 adjacent sides that are the same size and 3 angles which are 90. So the new structure has to be a square. And since it shares a side with the center square, it is the same size as it. So the area is made up of 5 such squares. Nov 17, 2017 at 17:02

Let $x$ be the area of the inner square, and $y$ be the area of each of the four coloured triangles shown in the picture below:

So we have: $x+4y=1$

Now each of these triangles has the same shape as $ADF$ which has hypotenuse $AF = \frac{\sqrt{5}}{2}$. The area of such a triangle is $\frac{h^2}{5}$ where $h$ is the hypotenuse, since this makes the area of the $ADF$ triangle exactly $\frac{1}{4}$ square centimeters. The hypotenuse of each coloured triangle is $1$, so we can now solve $y=\frac{h^2}{5}$ with $h=1$, and thus $y=\frac{1}{5}$, which gives the solution $x=1-4y=\frac{1}{5}$.

This relates to a proof of the Pythagorean theorem. One can see 4 right triangles surrounding the square. The largest side ('c') is AHD or 1 cm. One needs to find the lengths of the two shorter sides ('a' and 'b'), perhaps using proportional triangles. If found, then one can determine the inner square area is (a-b)*(a-b).

To find a and b, assume I is the point at the upper right corner of the inner square. AIF can be found by Pythagorean rule = root of 1*1 + (.5)*(.5) = root(5/4).

Triangles AID and DIF are congruent, so DI/IF = AI/DI, which means DIDI = AIIF. But AIF = root(5/4) so AI = root(5/4) - IF. So, DI*DI = root(5/4)IF - IFIF.

DIDI + IFIF = root(5/4)IF. But DIDI + IFIF = DFDF = .25. .25 = (root(5)/2)*IF or .5 = root(5)*IF or .5/root(5) = IF.

This means AI = root(5)/2 - .5/root(5) = 2/root(5). DI is thus 1/root(5) (since AD = 1).

In this case 'a' = AI = 2/root(5) and 'b' = DI = 1/root(5)

a-b is thus 2/root(5) - 1/root(5) = 1/root(5)

(a-b)*(a-b) = 1/5. The area of the square is 1/5.

Obviously I prefer @Michael Rozenberg's solution, as its straightforward to adapt for E, F, G, H not middle-points (clearly if AH=0 then the inner square is the outer square and the answer is 1; the posed question is AH=1/2 with answer 1/5; the other limit AH=AD=1 then clearly the answer is 0, but what of the intermediate values?).

So, to visually answer the question I'd go with a variation on the solution by @DeltaScuti_Fomalhautb. My large pink triangles are clearly each 1/4, so if subtracting them each from the main square (1-4* 1/4 =0), the leftover inner square is exactly what I've double-subtracted (the four darker shaded overlaps, one in each corner of the original square; ignore the extra stripey overlay).

Now to show that five of these double-counted triangles together make up my large pink triangles is obvious: The top-left pink triangle consists of the 'stripey-shaded' medium-sized triangle plus one of the overlappers, and the 'stripey-shaded' one is exactly four times the little overlapper it contains at it's right-hand point (it's congruent with the little one, and its hypothenuse is exactly twice the little one's hypothenuse, so its surface is four times the little one's). So in summary this proves the little square is four times the little overlap-triangle, each of which is 1/5 of the pink triangle which was 1/4 of the original square; so the little square is 4* 1/5 * 1/4 = 1/5.

Ignore the blue dotted lines, they'd give you another way to see it's five of those little triangles, but it's slightly more work to argue --- e.g., at this point I haven't established that the inner square's side is equal to the overlapper's base, and the dotted lines thus define shapes that seem congruent with the overlappers but might not be (like in that trick where you divide a rectangle into triangles but 1/16 or so gets lost as two lines aren't exactly parallel though drawing in fat pencil they seem; sorry can't find a link now).

• please take a look at my comment on Michael Rozenberg's answer. Nov 15, 2017 at 0:26
• Yes, that's exactly what I meant. Nov 15, 2017 at 17:20

I hope that the following makes sense. I would have been better if you had labelled points on the figure.

The side length of the square is $1$.

The area of one of the large triangles is $A_{\text{large}} = 1\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$.

The hypotenuse of the large triangle is $\sqrt{1^2 + \frac{1}{4}} = \sqrt{5}/2$.

One of the smaller triangles is congruent to the the larger triangle and it has hypotenuse $1/2$. The side lengths of the smaller triangle is therefore $$a= (\frac{1}{2} /\frac{\sqrt{5}}{2}) \cdot 1 = \frac{1}{\sqrt{5}} \\ b = (\frac{1}{2} /\frac{\sqrt{5}}{2}) \cdot \frac{1}{2} = \frac{1}{2\sqrt{5}}.$$ So the area of a smaller triangle is $$A_{\text{small}} = \frac{1}{20}.$$ Now you can probably figure out the rest. I think maybe I get the area of the smaller square to be $\frac{4}{20}$.

Let us borrow Inceptio's figure:

Since $ABX,\,ALY$ and $AZL$ are similar right-angled triangles (why?) and $B$ is the midpoint of $AL$, we conclude that $AX=XY$. For the similar reason, we have $AX=XY=LY=YJ=EJ=JF=XE=EC=\ell$ (say).

Furthermore, as $ABX$ and $ALY$ are similar right-angled triangles and $B$ is the midpoint of $AL$, we have $BX=\frac\ell2$. By similar reason, $BX=YZ=JK=DE=\frac\ell2$.

Therefore, if you glue the triangle $ABX$ and the trapezoid $BLXY$ together, you get a square with side length $\ell$. Similarly for the other three pairs of small triangles and trapezoids. As the square at the centre also has side length $\ell$, we see that its area is $\frac15$ the area of the large square.

• How do you begin by establishing that $\,ABX,ALY\,$ right-angled triangles? Mar 26, 2013 at 17:33
• @DonAntonio Angle BAX = angle DCE, and angle ABX + angle DCE is a right angle. Therefore angle AXB is a right angle. One also has to prove that angle AYL is a right angle, but I think the OP is capable of proving that. Mar 26, 2013 at 17:46
• Yup, the first part is right after establishing the congruence of the four big right-angled triangles, and all the rest follow in the same way. Mar 26, 2013 at 18:01

Let us happily borrow Inceptio's image:

Note that $$\,\Delta ALZ\cong\Delta LFK\cong\Delta FCD\cong \Delta CAB\,$$ as these all are right-angled triangles with legs of the same sizes $$\,1\,,\,0.5\,$$ .

The sum of these four triangle's areas is $$\,0.25\cdot 4=1\;\;(**)\,$$ , but note that the areas of the little triangles $$\,\Delta ABX\,,\,\Delta LZY\,,\,\Delta FKL\,,\,\Delta CED\,$$ are summed twice in the above calculation.

Now, we prove all these four little triangles are congruent to each other and right-angled. From the first congruence above we get:

$$\alpha:=\angle LAZ=\angle FLK=\angle DFC=\angle ACB$$

and, for example, in the triangle $$\,\Delta ALZ\,$$ , we get $$\,\beta:=\angle AZL=90^\circ-\alpha\,$$ , so in the triangle $$\,\Delta LZY\,$$ we have the pair of angles $$\,\alpha\,,\,\beta\,\,,\,\,\,\alpha+\beta=90^\circ\,$$ , and from here that this little triangle is right- angled. The same holds for the other three little triangles.

We also got the additional fact that $$\,XEJY\,$$ is a rectangle as its four angles are right, and in a little while more it will follow that it is actually a square.

From (**) above and the lines after it, it follows that the area of $$\,XEJY\,$$ equals the sum of the four little triangles.

We also get that the four trapezoids in the figure are congruent (as two of their sides are parts of the congruent little triangles and a third side equals the length of their hypotenuses, which is $$\,0.5\,$$ .

Now it is clear that $$\,XE=EJ=JY=YX\,$$ (since, for example, $$\,XE=BC-BX-EC=$$ (hypotenuse of big right angled triangle minus short lef of tine right-angled triangle-hypotenuse of tiny right angled triangle) and, as promised above, it follows $$\,XEJY\,$$ is in fact a square.

I now believe you can complete the exercise.

$$S_{\square_{Large}} = 4S_{\triangle_{Large}}$$ $$S_{\triangle_{Large}} = 5S_{\triangle_{Small}}$$ $$S_{\triangle_{Small}} = \frac14 S_{\square_{Small}}$$ $$\therefore S_{\square_{Large}} = 5 S_{\square_{Small}}$$