The Riddler, 2017-09-22 edition

After an extended 1+ month break, we’re back with a vengeance. Here’s a link to the problems. No code this week.

Riddler Express

We know the width of the top rectangle since we already have its area and length: \frac{24}{4} = 6. From here, we can determine the width of the lower rectangle by adding the two width differences: 6 + 3 + 2 = 11. This implies that the height of the bottom rectangle is \frac{44}{11} = 4, giving us our final answer of 4 inches1.

Riddler Classic

Let the height and width of the unknown rectangle be l and w, respectively. Now, we can derive two separate equations from the top-left and bottom-right rectangles:

(1)   \begin{align*} (11 - h) \times w &= 32 \\ (14 - w) \times h &= 45 \end{align*}

This is a system of equations with two variables and two unknowns which we can solve for using the quadratic equation (or, if you’re lazy like me, WolframAlpha):

(2)   \begin{align*} (h, w) &= (\frac{11}{2}, \frac{64}{11}) \\ (h, w) &= (\frac{45}{7}, 7) \end{align*}

The third rectangle will now allow us to determine which of the prior two is the correct solution. The first solution set (unknown area computes to \frac{11}{2} \times \frac{64}{11} = 32) gives us \frac{34}{11 - \frac{11}{2}} \approx 6.18 as the width of the upper-right rectangle, which is valid since \frac{64}{11} + 6.18 < 14. The second solution set (unknown area computes to \frac{45}{7} \times 7 = 45) gives us \frac{34}{11 - \frac{45}{7}} \approx 7.44 as the width, which is invalid since 7 + 7.44 > 14. Hence, the area of the unknown rectangle is 32 square inches.

1 Shame on you, FiveThirtyEight, for not using the metric system!