# 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.