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Locating Warehouse

Alignments to Content Standards: G-C.A.3 G-CO.D.12


You have been asked to place a warehouse so that it is an equal distance from the three roads indicated on the following map. Find this location and show your work.


  1. Show how to fold your paper to physically construct this point as an intersection of two creases.
  2. Explain why the above construction works, and in particular why you only needed to make two creases.

IM Commentary

This task can be implemented in a variety of ways. For a class with previous exposure to the incenter or angle bisectors, part (a) could be a quick exercise in geometric constructions,. Alternatively, this could be part of a full introduction to angle bisectors, culminating in a full proof that the three angle bisectors are concurrent, an essentially complete proof of which is found in the solution below.

We note also that the geometric construction aspect of the proof could be nicely replaced with an exploration involving the use of dynamic geometry software. Dragging the placement of the three initial roads in to a variety of configurations helps develop student's intuition for the incenter construction.


  1. Fold and crease the paper so that Oak lies on top of Rio. Do the same so that Oak lies on top of Elm. The point of intersection of the two creases is the point an equal distance from the three sides.

  2. Since the desired location should be an equal distance from three sides of triangle ABC, we are looking for the center of the circle inscribed in the triangle. The center of the inscribed circle, called the incenter, can be found by constructing the angle bisectors of the three interior angles of the triangle, as in the diagram below. Since these angle bisectors are concurrent, it is sufficient to construct two of the angle bisectors (and hence only make two creases in part (a)).


  3. Now we show the concurrence of the three angle bisectors: It is easy to see that the distance from the warehouse $W=WH$ to Rio equals the distance from $W$ to Oak. Namely, draw perpendiculars from $W$ to both Rio and Oak, with respective intersection points $X$ and $Y$. The triangles $\triangle WXC$ and $\triangle WYC$ are congruent since they are right triangles with $\angle WCX = \angle WCY$ and sharing side $WC$. So $WX=WY$. Similarly, drawing a perpendicular to Elm through $W$ meeting Elm at $Z$, we have $WY = WZ$. Combining the two equalities, we learn that $WX = WZ$, so that $W$ is on the angle bisector and the three angle bisectors are concurrent.