 # Problem of the Week Problem E and Solution Diagonal Distance

## Problem

Square $$ABCD$$ has $$K$$ on $$BC$$, $$L$$ on $$DC$$, $$M$$ on $$AD$$, and $$N$$ on $$AB$$ such that $$KLMN$$ forms a rectangle, $$\triangle AMN$$ and $$\triangle LKC$$ are congruent isosceles triangles, and also $$\triangle MDL$$ and $$\triangle BNK$$ are congruent isosceles triangles. If the total area of the four triangles is $$50$$ cm$$^2$$, what is the length of $$MK$$? ## Solution

Let $$x$$ represent the lengths of the equal sides of $$\triangle AMN$$ and $$\triangle LKC$$, and let $$y$$ represent the lengths of the equal sides of $$\triangle MDL$$ and $$\triangle BNK$$. Thus, area $$\triangle AMN =$$ area $$\triangle LKC= \frac{1}{2} x^2$$, and area $$\triangle MDL =$$ area $$\triangle BNK = \frac{1}{2} y^2$$.

Therefore, the total area of the four triangles is equal to $$\frac{1}{2} x^2+\frac{1}{2} x^2+\frac{1}{2} y^2+\frac{1}{2} y^2= x^2 + y^2$$. Since we’re given that this area is $$50$$ cm$$^2$$, we have $$x^2 + y^2 = 50$$.

Three different solutions to find the length of $$MK$$ are provided.

Solution 1

In $$\triangle AMN$$, $$MN^2=AM^2+AN^2 = x^2+ x^2$$, and in $$\triangle BNK$$, $$NK^2=BN^2+BK^2= y^2+y^2$$.

Since $$MK$$ is a diagonal of rectangle $$KLMN$$, then by the Pythagorean Theorem we have \begin{aligned} MK^2 &=MN^2+NK^2\\ &=x^2+x^2+y^2+y^2\\ &=x^2+y^2+x^2+y^2\\ &=50+50\\ &=100 \end{aligned} Since $$MK>0$$, we have $$MK= 10$$ cm.

Solution 2

In $$\triangle AMN$$, $$MN^2= x^2 + x^2=2x^2$$. Therefore, $$MN=\sqrt{2}x$$, since $$x >0$$.
In $$\triangle BNK$$, $$NK^2= y^2 + y^2= 2y^2$$. Therefore, $$NK=\sqrt{2}y$$, since $$y > 0$$.

Since $$MK$$ is a diagonal of rectangle $$KLMN$$, then by the Pythagorean Theorem we have \begin{aligned} MK^2&=MN^2+NK^2\\ &=(\sqrt{2}x)^2+(\sqrt{2}y)^2\\ &=2x^2+2y^2\\ &=2(x^2+y^2)\\ &=2(50)\\ &=100 \end{aligned} Since $$MK>0$$, we have $$MK= 10$$ cm.

Solution 3

We construct the line segment $$KP$$, where $$P$$ lies on $$AD$$ such that $$KP$$ is perpendicular to $$AD$$.

Then $$APKB$$ is a rectangle. Furthermore, $$AP = BK = y$$, $$PK = AB = x + y$$, and $$PM = AM - AP = x - y$$. Since $$\triangle PKM$$ is a right-angled triangle, by the Pythagorean Theorem we have \begin{aligned} MK^2&=PM^2+PK^2\\ &=(x-y)^2+(x+y)^2\\ &=x^2 - 2xy + y^2 + x^2 + 2xy + y^2\\ &=2x^2+2y^2\\ &=2(x^2+y^2)\\ &=2(50)\\ &=100 \end{aligned} Since $$MK>0$$, we have $$MK= 10$$ cm.