I have always loved math, and I don’t think there’s any field within math I love more than algebra. I certainly enjoyed calculus, and there are parts of number theory that still fascinate me (Goldbach’s conjecture, an unproven hypothesis that dates back to 1742, more than anything else), but algebra just speaks to me like nothing else in the domains of math or science. So when I saw this week’s FiveThirtyEight Riddler problem, which boils down to solving two equations for two unknowns, I might have dropped everything I was doing and spent about a half an hour solving it – messily, but I think ultimately getting the right answer.
My own personal love of algebra dates back to when I was ten years old, and my junior high school, lacking an honors math class for sixth grade, decided instead to bump me up to the regular eighth grade math course, where I met a wonderful teacher and some awful kids. (I was three years younger than they were, and of course small for my age anyway.) I was told very little about the move but somehow understood that I’d be learning algebra, so I went to the school library and found a book, long out of print now, called Realm of Algebra, by an author I’d never heard of at the time but would later grow to know very well through his science fiction writing, Isaac Asimov. I devoured the book, which I credit to Asimov’s ability to make even abstruse concepts clear to readers, in a weekend, and ended up ahead of where I needed to be for the class. Algebra felt to me like another language, as easy to comprehend as English, maybe even more so – like this was my native tongue and everyone had been hiding it from me. I could always “think” in numbers, but algebra gave me an entire framework for it, and everything I learned that year, especially from Asimov’s book, still directs much of how I think about problems today.
It has also made me an easy mark for puzzles and games that revolve around algebraic questions. I often check FiveThirtyEight’s Riddler questions, but I rarely try to solve them – some look too hard or involved, some just don’t grab me. This week’s question, about finding the area of a missing rectangle, hooked me from the start. (I suppose I should disclose that FiveThirtyEight is part of the ESPN network of sites, and thus I am connected to it as well.) Here’s the question in brief: Find the missing area in the picture below, bearing in mind that it is not to scale.
I tweeted the link to the article last night and got a slew of responses from readers, some right, some I don’t think were right, and a few that gave me more insight into the problem – one of which made me realize my first answer was impossible – so I decided to take a few minutes and explain my method, in case it’s useful to anyone or still contains a mistake.
To figure out the area of the lower-left rectangle, you need to know its height and width, so this is a problem of two unknowns, which means you need (at least) two equations containing those unknowns to be able to solve it. To make the math a little less messy, I defined the height (vertical axis) of the target area as 11-x and the width (horizontal axis) as 14-y, meaning that the height of the upper-left area is x and the width of the lower-right area is y. We know the area of the lower right rectangle is 45, and the area of the upper left rectangle is 32, so using the formula for the area of a rectangle we get the following two equations:
(14 – y)x = 14x – xy = 32
y(11 – x) = 11y – xy = 45
You can then solve one equation for one variable in terms of the other, substitute that back into the second equation, and solve for one of the two variables. I chose to solve the first equation for x, yielding:
x(14 – y) = 32
x = 32/(14 – y)
11y – y(32/(14 – y)) = 45
11y – 32y/(14 – y) = 45
11y(14 – y) – 32y = 45(14 – y)
154y – 11y2 – 32y = 630 – 45y
–11y2 + 167y – 630 = 0
That, my friends, is a quadratic equation, and if you remember your quadratic formula – where you take the two coefficients and the constant and plug them into the formula to get the two possible solutions – you can solve it from here, or you can just plug them into this site and get your two answers for y, which in this case are 7 or 90/11.
It turns out that both answers produce whole-number, positive results for the area of the lower left rectangle. If y is 7, x is 32/7, and the area is 45. If y is 90/11, x is 11/2 (5.5) and the area is 32. A reader pointed out that the first answer is impossible, however, because of the one number I haven’t mentioned yet: the truncated, upper-right rectangle’s area, which is 34. Because the area of the whole shebang has to be less than 154 (11 * 14), since there’s a piece missing at the extreme upper right, then the lower left area has to be less than (154 – 32 – 34 – 45), which is 43. That leaves 32 as the only possible answer. (EDIT: Fixed the second sentence, where I transposed the two answers.)
I think.
Is it weird that I was annoyed at the claim that the only math you need to know is the area of a rectangle? Because I wasted some time thinking I must be doing it wrong, since I couldn’t figure how to solve it without looking up the quadratic formula.
Funny…I did it the opposite way of you (but with the same logic) and got the same answer.
x*y = 32
(11-x)*(14-y) = 45
154-11y-14x+xy = 45
154-11y-14x+32 = 45
186-11y-14x = 45
11y+14x = 141
352/x+14x = 141
14x^2 – 141x + 352 = 0
x is 11/2 or 32/7
so, y is 64/11 or 7, respectively
our solutions are either 11/2*64/11 = 32 or 45/7*7 = 45
Your method is much cleaner than mine, but I tackled it by determining the area of the “missing rectangle” in the upper right corner. Using that, I could solve for “?” through subtraction from the total area. The benefit to that is that the quadratic equation to solve for the area of the “missing rectangle” only yielded one positive root, not two. Anyway, I came up with 32 as well.
The missing area had to be 11, and then the total area was 154, so you get 154-32-34-11-45=32, right? I thought, from their disclaimer, that doing so used an unacceptable assumption about the image, that all four visible rectangles meet at a single corner in the center. It turns out that they do, so your method ends up correct.
Appreciate the write up, though I am irrationally annoyed at your use of “x” for the vertical variable and “y” for the horizontal variable.
That’s funny. It didn’t even occur to me that I’d reversed the standard use of the two.
“A reader”!
was it you? I had so many replies this morning that I couldn’t find the tweet.
But if y=7 and x=4.571429, doesn’t that screw up the top left rectangle? Wouldn’t that be 7*6.428571 and then *not* be 32 anymore?
https://twitter.com/cjvk/status/912321196208693251
That’s the one! Thank you. I woke up with the problem still on my mind, like I’d screwed up but had no idea why. I think I’d failed to run both solutions through and that’s how I missed it. Your tweet showed me my mistake.
Could it be that the trick is the top two rectangles are longer on the x axis than the diagram shows, since it “isn’t to scale?
That would allow for 45 being the area of the unknown. With 7 being the unknown x axis piece and 6.428571 being the missing y axis piece. Then the top 32 works, and the top 34 actually hangs over the bottom 45 since the total top length would have to be 14.4375.
I went to the web page they linked for the book the puzzle was taken from. It says this about all the puzzles:
“Area = length × width
Use spatial reasoning to find helpful relationships
Whole numbers are all you need. You can always get the answer without using fractions!”
That tells me that there’s a simple solution that does not require remembering how to do quadratic equations. I haven’t figured it out yet though.
It has to be that you assume the line segments that split the area into four rectangles are both straight and continue through their intersection, so that you take the total area (14*11=154), then subtract what’s known … I have a problem with this, given the instructions that the drawing isn’t to scale, which says to me that you shouldn’t make that assumption.
The description specifically states that the four rectangles meet at a single point in the center, so being rectangles (as opposed to other quadrilaterals) they necessarily must continue straight through their intersection.
The problem with subtracting what’s known is that there are still two unknown areas within the 11×14=154 area. No matter how I try to look at it, there always seem to be at least two unknowns, which means using math more complicated than Area = length x width.
My breakdown so far:
* Four unknown lengths along the x axis: a + b + c = 14 for the upper rectangles and a + d = 14 for the lower ones.
* Two unknown lengths on the y axis: e + f = 11.
* Two unknown areas: 32 + 34 + 45 + m + n = 154.
We know a*e = 32, b*e = 34, and d*f = 45, and area m (the “?” we’re solving for) = a*f, and area n = e*c. As far as I can tell, that is everything we know about the dimensions and areas of everything in play.
The description of the puzzle suggests that there’s a simple solution than yours to finding “?”, but I can’t see it.
“If y is 7, x is 32/7, and the area is 32. If y is 90/11, x is 11/2 (5.5) and the area is 45.”
This is not correct – and explains why your answer is “opposite” of mine. (I figured out 32 but concluded that x=5.5, incidentally using the same x-as-upper-left-height).
Correct: If y is 7, x is 32/7, and the area is (11-32/7)*(14-7) = [(77-32)/7]*7 = 45. If y is 90/11, x is 5.5, and the area is 32.
And also of course “A reader pointed out that the second answer is impossible” becomes “A reader pointed out that the first answer is impossible”.
so, the answer is 45, right?
No it’s 32.
Thanks, Chris. I fixed the post. I tried to write this up from the handwritten solution I’d crafted last night, and screwed it up.
I did the problem the quadratic equation way firstly, but the eureka moment came afterwards. The area of the two rectangles kitty corner to each other (32 + 45 = 77) is half of the area of the entire bounding rectangle (11 x 14 = 154). Because of this, it seems like the unknown area must be equal to the area of one of the adjacent rectangles by necessity, though I don’t know how to prove that. But it makes some intuitive sense, to me at least.
On a side note, Asimov was brilliant in that ability. He made math accessible to you; he made history a passion of mine. Now I’m a history teacher, largely because of The Dark Ages, The Roman Empire, and The Middle East: 10,000 Years of history.
Great post. I think as closer than I thought. Had the first non-possible answer and knew it was wrong.
I had 6 variables though!
Doing the express puzzle clued you in a little that the line went straight through.
Andre – Brilliant! That may be the key I was looking for. I just sketched out some example rectangles and it appears as though it’s only possible for those two sections to be exactly half the area of the rectangle if one of the lines that splits the rectangle intersects it at the exact halfway point along its axis. So in this case, the horizontal line that divides the larger rectangle into upper and lower sections must intersect with the y axis at the exact halfway point, which makes the upper left and lower left rectangles exactly the same – 32.
I don’t know and can’t prove that that’s an actual property of rectangles though, so I’m not completely comfortable with that answer.
But this is not enough. While it is the case that ? = 32, and that the horizontal line divides evenly (and that these two are related), what is it about the numbers 32, 45, 11, and 14 that can lead one to the conclusion that the horizontal line must be in the middle?
Chris – That’s my problem. I can’t prove that. All I’ve done is try to find another set of rectangles where it’s not true. Here’s what I’ve found by crude trial and error:
|—-|——–|
| a | b |
|—-|——–|
| c | d |
|—-|——–|
If area of a+d or b+c = 1/2 total area, then one of the intersecting lines is at the halfway point. I can’t logically prove why that must be so (or even if it must be so), but I also can’t find an example where it isn’t true. That’s why I’m not satisfied with that approach.
What is funny is that once you do the two equations, two unknowns – the quadratic equation yields 11/2 for the vertical length. Given that the overall length is 11, this means that the two stacked rectangles have the same height. You don’t actually need to do anything else. The rectangles HAVE to be the same area, 32.
It looks like the answer is now posted, but I swear I didn’t look before finding this solution 🙂
—–
—–
—–
1) Call the upper left rectangle 1, the upper right rectangle 2, the lower right rectangle 3, the lower left rectangle 4, and the missing space in the upper right 5. Define Ai to be the area of rectangle i, and define A = 11*14 = 154 to be the total area of all five rectangles. Thus, A = A1+A2+A3+A4+A5.
2) A1+A3 = 32+45 = 77 = A/2. Therefore, A1+A3 = A/2 = A2+A4+A5.
3) Define the width and height of rectangle i to be xi and yi, respectively. Thus, Ai = xi*yi. Also note that x4 = x1 and y4 = y3.
4) From 3) we can see that A2+A5 = x3*y1, and A4 = x1*y3. Plugging this information into 2) gives us x1*y1 + x3*y3 = x1*y3 + x3*y1.
5) Rearranging and grouping terms in 4) gives (x1 – x3)*(y1 – y3) = 0. Therefore, x1 = x3 and/or y1 = y3.
6) We know that A2 > A1 means that x2*y2 > x1*y1. Because y1 = y2, we see that x2 > x1. We also see from the image that x3 > x2. Therefore, x3 > x1, meaning that x3 != x1. To satisfy 5) this means we have to have y1 = y3.
7) Since y1 = y3 = y4 and x1 = x4, we can conclude that x1*y1 = x4*y4 —-> A4 = A1 = 32.
As mentioned before, the 45 and 32 rectangles equal half the area of the completed big rectangle. Since there are vertical and horizontal dissecting lines and two of the four resulting rectangles equal half of the area, can’t you infer two sets of congruent rectangles? Since 34 is greater than 32, it implies the area of the “?” rectangle is 32.
I think you can, Dustin. The solution I posted appears to be what you said, but done explicitly (and much less succinctly!).
Well, looks like I was on the right track and all I was missing was being able to prove that my hunch about the horizontal axis splitting the full rectangle in two was correct.