Squares $ABCD, DCGH, BEFG$ and $ELKM$ are positioned as shown on the picture. Find the area of triangle $DGK$ if you know that the area of square $ABCD$ is $20$.

enter image description here


Here's another solution using the fact that $\triangle ABC$ and $\triangle ABD$ have the same area if $AB \parallel CD$:

enter image description here

The reason is that $CE = DF$, so $\frac12 AB \cdot CE = \frac12 AB \cdot DF$.

Now, since $DG \parallel EK$ and $BD \parallel EG$, we have $\triangle DGK = \triangle DEG = \triangle BEG$ in area:

enter image description here

The result is hence

$2 \cdot 20 = 40$.

  • $\begingroup$ Yes, well this is basicly the same solution as hexomino's. $\endgroup$
    – Greedoid
    Apr 17 at 6:15
  • 1
    $\begingroup$ @Greedoid It may use the same principle as Hex's proof, but it applies that principle in a different manner resulting in different triangles. $\endgroup$ Apr 17 at 7:20

Here is a quick way to do it

The locus of the point $K$ as the side length of the square $EMKL$ varies is a line which is parallel to $DG$. Hence, if we consider $DG$ as the base of the triangle $DGK$ then its perpendicular height, and thus its area, is invariant with respect to the size of the square $EMKL$.

This means that we could perform the calculation with the side length $EM$ being anything we like and the answer would be the same. If we make it so that the point $M$ coincides with the point $F$, it is easy to compute that the area of triangle $DGK$ is twice the area of the square $ABCD$ and thus the answer is $40$.

  • $\begingroup$ I like this puzzle - at first sight there is a constraint missing but the answer comes out cleanly :) I think this puzzle would be suitable for some math channel such as 3Blue1Brown $\endgroup$
    – happystar
    Apr 17 at 0:01
  • $\begingroup$ Not worth a separate answer but I'd say the most opportunistic placement of M would be as the midpoint of E and F. $\endgroup$
    – loopy walt
    Apr 17 at 7:37

Well, this answer is a bit late, but still quite simple I believe:

Let's introduce a coordinate system with origin at $A$, having $AL$ and $AH$ as $x$ and $y$ axes respectively. Also, let $AB=1$ (well, the $ABCD$ square has now an area of $1$, not $20$, but we can just scale the final result 20 times, i.e. assuming that we scaled down the entire picture $\sqrt{20}$ times on each axis - otherwise we had to deal with a bunch of square roots). Now, $D$ has coordinates $(0, 1)$, $G$ is $(1,2)$ (since the squares $ABCD$ and $DCGH$ are equal with side $1$), and $K$ is $(s + 3, s)$ where $s$ is the side of $ELKM$ square, which we do not know. (The $BEFG$ square must have the side of $2$, because $BG=BC+CG=1+1=2$, so $E$ is at $(3, 0)$).
Now calculate the area of $DGK$ triangle from the coordinates of its vertices using the well-known formula: $$A=\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|,$$ where $x_i$ and $y_i$ are the coordinates of $i$-th vertex ($i=1,2,3$).
Plugging $x_1=0, y_1=1, x_2=1, y_2=2, x_3=s+3, y_3=s$ gives $$A=\frac12|0(2-s)+1(s-1)+(s+3)(1-2)|=\frac12|s-1-s-3|=\frac12\times4=2.$$ Thus, the final result (which does not depends on $s$) is $2\times20=40$.


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