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A Midpoint Miracle


Alignments to Content Standards: G-GPE.B.4 G-GPE.B.5

Task

Draw a quadrilateral $ABCD$. Try to draw your quadrilateral so that no two sides are congruent, no two angles are congruent, and no two sides are parallel.

  1. Let $P$, $Q$, $R$, and $S$ be the midpoints of sides $AB$, $BC$, $CD$, and $DA$, respectively. Use a ruler to locate these points as precisely as you can, and join them to form a new quadrilateral $PQRS$. What do you notice about the quadrilateral $PQRS$?
  2. Suppose your quadrilateral $ABCD$ lies in the coordinate plane. Let $(x_1, y_1)$ be the coordinates of vertex $A$, $(x_2, y_2)$ the coordinates of $B$, $(x_3, y_3)$ the coordinates of $C$, and $(x_4, y_4)$ the coordinates of $D$. Use coordinates to prove the observation you made in part (a).

IM Commentary

This classroom task gives students the opportunity to prove a surprising fact about quadrilaterals: that if we join the midpoints of an arbitrary quadrilateral to form a new quadrilateral, then the new quadrilateral is a parallelogram, even if the original quadrilateral was not.

Student observations about the quadrilateral $PQRS$ in part (a) will vary. Some students may conjecture that the quadrilateral is a rhombus or rectangle, depending on how they drew the original quadrilateral. Teachers may find it helpful to have several students post the pictures they drew in part (a) in order to convince the class that while $PQRS$ always seems to be a parallelogram, it may not be a rhombus or rectangle.

One possible variant on part (a) is to have students draw the quadrilateral in GeoGebra, assuming that they have computers available. This allows students to move various parts of the quadrilateral around and see that the midpoint quadrilateral remains a parallelogram no matter where the parts are moved. This strategy also allows students to see that in general, the midpoint quadrilateral is not a rhombus or rectangle.

This fact provides an elegant example of a theorem in geometry that is somewhat tricky to prove using classical methods, but relatively easy to prove using coordinates. Students working on this task will have to use the formula for the midpoint of a line segment and the criterion for whether two lines are parallel (G-GPE.5).

Solutions

Solution: Solution

  1. Midpointmiraclebare_bfb95d4b66c46114fc00dcaefb9a4f78

    The quadrilateral PQRS appears to be a parallelogram. (In fact, in the picture above, it appears to be a rhombus, but if we draw a quadrilateral ABCD with two consecutive sides much shorter than the other two sides, we can see that the sides of PQRS need not be equal.)
  2. To prove that PQRS is a parallelogram, we will check that side PQ is parallel to SR and that QR is parallel to PS.

    Midpointmiracle_99a28f838997ce9518fc0cae0a741622

    With the coordinates of A, B, C, and D as given above, we have \begin{eqnarray*} P & = & \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \\ Q & = & \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) \\ R & = & \left( \frac{x_3 + x_4}{2}, \frac{y_3 + y_4}{2} \right) \\ S & = & \left( \frac{x_4 + x_1}{2}, \frac{y_4 + y_1}{2} \right) \\ \end{eqnarray*} So the slope of the line PQ is \begin{eqnarray*} \frac{\Delta y}{\Delta x} & = & \frac{\frac{y_2 + y_3}{2} - \frac{y_1 + y_2}{2}}{\frac{x_2 + x_3}{2} - \frac{x_1 + x_2}{2}} \\ & = & \frac{(y_2 + y_3) - (y_1 + y_2)}{(x_2 + x_3) - (x_1 + x_2)} \\ & = & \frac{y_3 - y_1}{x_3 - x_1} \end{eqnarray*} The slope of the line SR is \begin{eqnarray*} \frac{\Delta y}{\Delta x} & = & \frac{\frac{y_3 + y_4}{2} - \frac{y_4 + y_1}{2}}{\frac{x_3 + x_4}{2} - \frac{x_4 + x_1}{2}} \\ & = & \frac{(y_3 + y_4) - (y_4 + y_1)}{(x_3 + x_4) - (x_4 + x_1)} \\ & = & \frac{y_3 - y_1}{x_3 - x_1} \end{eqnarray*} These slopes are the same; therefore, the lines PQ and SR are parallel. We can compute the slopes of lines QR and PS to find that both slopes are equal to $$ \frac{y_4 - y_2}{x_4 - x_2}. $$ Therefore, QR and PS are parallel as well. Since the quadrilateral PQRS has two pairs of parallel sides, it is a parallelogram.

Solution: Alternative Solution to (b)

We may simplify the algebraic work of this proof slightly by applying a transformation to the quadrilateral ABCD before we begin. Suppose that, by applying a translation and a rotation as necessary, we move the quadrilateral ABCD so that the vertex A is at the origin $(0, 0)$, and the vertex B is somewhere on the x-axis, at the point $(x_2, 0)$. We define $(x_3, y_3)$ and $(x_4, y_4)$ to be the new coordinates of the vertices C and D, respectively.

We now have \begin{eqnarray*} P & = & \left( \frac{x_2}{2}, 0 \right) \\ Q & = & \left( \frac{x_2 + x_3}{2}, \frac{y_3}{2} \right) \\ R & = & \left( \frac{x_3 + x_4}{2}, \frac{y_3 + y_4}{2} \right) \\ S & = & \left( \frac{x_4}{2}, \frac{y_4}{2} \right) \\ \end{eqnarray*} So the slope of the line PQ is \begin{eqnarray*} \frac{\Delta y}{\Delta x} & = & \frac{\frac{y_3}{2} - 0}{\frac{x_2 + x_3}{2} - \frac{x_2}{2}} \\ & = & \frac{y_3 - 0}{(x_2 + x_3) - x_2} \\ & = & \frac{y_3}{x_3} \end{eqnarray*} The slope of the line SR is \begin{eqnarray*} \frac{\Delta y}{\Delta x} & = & \frac{\frac{y_3 + y_4}{2} - \frac{y_4}{2}}{\frac{x_3 + x_4}{2} - \frac{x_4}{2}} \\ & = & \frac{(y_3 + y_4) - y_4}{(x_3 + x_4) - x_4} \\ & = & \frac{y_3}{x_3} \end{eqnarray*} These slopes are the same, so the lines PQ and SR are parallel. We can compute the slopes of lines QR and PS to find that both slopes are equal to $$ \frac{y_4}{x_4 - x_2}. $$ So QR and PS are parallel as well. Since the quadrilateral PQRS has two pairs of parallel sides, it is a parallelogram.

jharmon says:

almost 5 years

Should the opening sentence read "Draw a quadrilateral ABCD. Try to draw your quadrilateral ..."?

Ashli says:

almost 5 years

Agreed. Especially since the statement wants no two sides to be parallel. Thanks!

Monique Rousselle Maynard says:

almost 5 years

As we look at this task, we wonder about how different samples of student work would be assessed along the demonstration of proficiency continuum. For example, if a Geometry student who struggles with algebra were able to demonstrate the property in a concrete manner on the coordinate plane, would that be sufficient to demonstrate proficiency on this standard or must that student at least approach the generalization? In other words, if a student does not even try to generalize the property, would that student be limited to "nearly meets proficiency" on this question?

stickler says:

almost 5 years

@Monique: Standard G.GPE.4 in its wording indicates that "prove or disprove that the point (1, sqrt(3)) lies on the circle centered at the origin and containing the point (0, 2)" is an example of what it means to prove "simple geometric theorems algebraically" -- so I think we can presume that proving something is true for specific figures (i.e. not general cases) is allowed by the standard. Indeed, one possible interpretation of the word "simple theorems" in the standard is simply proving specific statements about specific figures. Obviously, being able to write an algebraic proof that is completely general is valuable, and certainly the standard does not rule that out. However, I do not think the standard explicitly implies all "algebraic proofs" must be general ones.

I suspect that assessment tasks that will be computer-scored may be likelier to posit specific givens, rather than general ones, if only because it is easier to imagine a computer-scorable item that uses specific values. A performance-based assessment item that really seeks to have students make a generalized argument would likely require a constructed response to be scored by humans.

Cam says:

almost 5 years

Hi Monique, thanks for your comments.

I'd concur that this task, like most other tasks that begin with making open-ended observations, serves better in an instructional setting than an assessment one. For example, an an instructional setting, if by "demonstrate the property in a concrete manner on the coordinate plane" you mean a student was able to furnish a complete calculation for one specific parallelogram, then I would argue that this was certainly progress toward the standard and that the student should be encouraged to develop their algebraic skills trying to mimic the argument with an arbitrary parallelogram.

So while I think the original intent of the task is to emphasize the "prove simple geometric theorems algebraically," thereby necessitating the need to approach the generalization, reasonable modifications to the task fitting it to a more assessive setting are not hard to imagine: Students could be given a fixed quadrilateral, or asked to construct one themselves, and then asked to demonstrate the parallelogram-midpoint property for a single quadrilateral.

Richard Askey says:

almost 5 years

A synthetic proof is so nice it is a shame not to illustrate it, and suggest how it can be suggested if a student does not find it before doing a coordinate proof. The first of the coordinate proofs is much nicer than the second since symmetry has not been broken by the special choice of points for two vertices. After the first proof has been found, ask students to see if there is another way of finding a line with the same slope that PQ has. Some will almost surely observe that AC has the same slope, and so is parallel to PQ. They should have learned that AC and PQ are parallel from what they have done with similarity of triangles. Then the same arguments apply to AC and SR, and BD and the other two segments connecting midpoints.

Cam says:

almost 5 years

Thanks for the comment, Richard. We agree completely, and have in the works a variation of this task that takes the synthetic point of view. We'll then have each task reference the other, so that instructors have the option to address either/both approach.