#
Problem
of the Week

Problem
D and Solution

Where
is Pete?

## Problem

Amir, Bita, Colin, and Delilah are standing on the four corners of a
rectangular field, with Amir and Colin at opposite corners. Pete is
standing inside the field \(5\) m from
Amir, \(11\) m from Bita, and \(10\) m from Delilah. In the diagram, the
locations of Amir, Bita, Colin, Delilah, and Pete are marked with \(A,~B,~C,~D,\) and \(P\), respectively.

Determine the distance from Pete to Colin.

Hint: Consider drawing a line segment
through \(P\), perpendicular to two of
the sides of the rectangle, and then using the Pythagorean Theorem.

## Solution

We start by drawing a line through \(P\), perpendicular to \(AB\) and \(DC\). Let \(Q\) be the point of intersection of the
perpendicular with \(AB\) and \(R\) be the point of intersection with \(DC\).

Since \(QP\) is perpendicular to
\(AB\), \(\angle AQP = 90^{\circ}\) and \(\angle BQP = 90^{\circ}\). Since \(PR\) is perpendicular to \(DC\), \(\angle
DRP = 90^{\circ}\) and \(\angle CRP =
90^{\circ}\). We also have that \(AQ =
DR\) and \(BQ = CR\).

We can apply the Pythagorean Theorem in \(\triangle AQP\) and \(\triangle BQP\).

From \(\triangle AQP\), we have
\(AQ^2 + QP^2 = AP^2= 5^2 = 25\).
Rearranging, we have \[QP^2 = 25 - AQ^2
\tag{1}\]

From \(\triangle BQP\), we have
\(BQ^2 + QP^2 = BP^2 = 11^2 = 121\).
Rearranging, we have \[QP^2 = 121 - BQ^2
\tag{2}\]

Since \(QP^2 = QP^2\), from \((1)\) and \((2)\) we find that \(25 - AQ^2 = 121 - BQ^2\) or \(BQ^2 - AQ^2 = 96\). Since \(AQ = DR\) and \(BQ = CR\), this also tells us \[CR^2 - DR^2 = 96 \tag{3}\]

We can now apply the Pythagorean Theorem in \(\triangle DRP\) and \(\triangle CRP\). From \(\triangle DRP\), we have \(DR^2 + RP^2 = DP^2= 10^2 = 100\).
Rearranging, we have \[RP^2 = 100 - DR^2
\tag{4}\]

When we apply the Pythagorean Theorem to \(\triangle CRP\) we have \(CR^2 + RP^2 = CP^2\). Rearranging, we have
\[RP^2 = CP^2 - CR^2 \tag{5}\]

Since \(RP^2 = RP^2\), from \((4)\) and \((5)\) we find that \(100 - DR^2 = CP^2 - CR^2\), or \[CR^2 - DR^2 = CP^2 - 100 \tag{6}\]

From \((3)\), we have \(CR^2 - DR^2 = 96\), so \((6)\) becomes \(96=CP^2 - 100\) or \(CP^2 = 196\). Thus \(CP = 14\), since \(CP > 0\).

Therefore the distance from Pete to Colin is \(14\) m.