# Silver Rectangle

Alignments to Content Standards: A-CED.A.2

Below is a picture of a rectangle $ABCD$ with segment $\overline{MN}$ drawn where $M$ is the midpoint of $\overline{BC}$ and $N$ is the midpoint of $\overline{AD}$: Suppose $ABCD$ is similar to $BMNA$. What is $\frac{|BC|}{|AB|}$?

## IM Commentary

This task provides a geometric context for working with ratios and algebraic equations. Students will create and then solve an algebraic equation describing a remarkable shape, the silver rectangle. This rectangle is characterized by being similar to the two smaller rectangles obtained when it is cut in half (along the longer sides). The similarity hypothesis leads to an equation in terms of side lengths of the different rectangles. Using this equation and the fact that $M$ and $N$ are the midpoints of their respective sides leads to a quadratic equation which students then solve. A little algebraic manipulation gives the desired quotient of side lengths.

Students will apply prior knowledge of ratios to set up an equation. The teacher may wish to recall beforehand the proportional equations which arise when two geometric figures are similar, perhaps in the case of two triangles. The activity is appropriate for group work and the teacher may need to be proactive helping students set up ratios from the given similarity. In addition, the similarity of the two rectangles gives rise to many equivalent ratios and the teacher may wish to encourage students to choose these so as to involve the two side lengths in question, $|BC|$ and $|AB|$. The work required to proceed from the given information to this algebraic equation exemplifies MP2, Reason Abstractly and Quantitatively.

In the solution provided, the ''variables'' which appear are not the usual ones represented by letters like $x$ and $y$. They are the unknown side lengths of the line segments from the picture. If students prefer, they can label the different side lengths of the rectangle with letters. For example we could set $y = |BC|$ and $x = |AB|$. Then the method of the solution leads to the equation $$\frac{y^2}{2} = x^2.$$ This gives $y = \sqrt{2}x$ and so the quotient $\frac{y}{x}$ is again $\sqrt{2}$. If both methods of solution come up in class, the teacher may wish to have students share and understand the relationship between the two.

Instructions and an analysis of how to produce a silver rectangle with origami is presented here: www.illustrativemathematics.org/tasks/1488.

## Solution

We know by hypothesis that rectangle $ABCD$ is similar to rectangle $BMNA$. Since corresponding parts of similar geometric figures are proportional we obtain $$\frac{|BC|}{|AB|} = \frac{|MN|}{|BM|}.$$ We also know that $|BC| = 2|BM|$ since $M$ is the midpoint of $\overline{BC}$. Plugging this into the equation above and cross multiplying we find $$2|BM|^2 = |AB||MN|.$$ We know that $|MN| = |AB|$ since $BMNA$ is a rectangle so this gives $$2|BM|^2 = |AB|^2.$$ Solving this equation we find (using the fact that side lengths must be positive, excluding the negative solution) $$|BM| = \frac{\sqrt{2}|AB|}{2}.$$ We know that $|BC| = 2|BM|$, since $M$ is the midpoint of $\overline{BC}$, so this means that $|BC| = \sqrt{2}|AB|$. We can now answer the question: $$\frac{|BC|}{|AB|} = \sqrt{2}.$$