#
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.