CEMC Banner

Problem of the Week
Problem D and Solution
Where is Pete?


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.

Rectangle ABCD with point P in its interior. AP is 5 units
long, BP is 11 units long, and DP is 10 units long.

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.


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.