A square is divided into 8 rectangles whose areas are equal. The height of one rectangle is shown as 8 cm.

What is the area of the entire square?

As usual, watch the video for a solution.

Area of Square From 8 Rectangles Equal Area

Or keep reading.

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Answer To Area of Square From 8 Rectangles Equal Area

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The answer is 1225 cm². Here is how to derive it. Label the rectangles from B to I as shown, and let a be the width of rectangle B.

The area of B must be equal to 8a. All rectangles have the same area, so all have an area of 8a.

Consider the rectangle formed by B, C, and D. The total area is then 3(8a) = 24a. Since this rectangle width is also a, its height must be 24. That is the height of rectangle E.

Now iterate these same steps by joining rectangle E, which adds an area of 8a, to the previous rectangle. The new rectangle’s area is 32a, and its height is 24. So the width must be 32a/24 = 4a/3. This is the width of rectangle F.

Now append the rectangle F to increase the area by 8a. The new rectangle’s area is 40a, and its width is 4a/3. So the height must be 40a/(4a/3) = 30. This will be the height of rectangle G.

Next append the rectangle G to increase the area by 8a. The new rectangle’s area is 48a, and its height is 30. So the width must be 48a/30 = 8a/5. This will be the width of rectangle H.

Then append the rectangle H to increase the area by 8a. The new rectangle’s area is 56a, and its width is 8a/5. So the height must be 56a/(8a/5) = 35. This will be the height of rectangle I.

But the height of rectangle I is exactly the side of the square. So the square’s area is then 35² = 1225 cm².

For good measure we can easily solve for a. Append I to increase the area to 64a. But we know this equals 1225, so we must have a = 1225/64 cm.

From there one could solve for the side lengths of all the rectangles, if so desired.

What a fun problem!

References

Scientific American
https://www.scientificamerican.com/game/math-puzzle-great-lengths/

Previous puzzle 5 rectangles
https://www.youtube.com/watch?v=lrND8AeL23s
https://mindyourdecisions.com/blog/2020/12/20/5-rectangles-puzzle/

As usual, watch the video for a solution.

Square In A 3-4-5 Triangle Puzzle

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Answer To Square In 3-4-5 Triangle

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

We will first show that triangles ABC, GCF, FEB, and ADG are similar.

Triangles ABC and GCF share a right angle at C. Since GF is paralell to DE it is also parallel to AB, and so CGF is a corresponding angle to CAB and CFG is a corresponding angle to CBA. Thus ABC and GCF are similar triangles.

Since DEFG is a square, triangles FEB and ADG are right triangles, but each also shares one acute angle with ACB, and so these triangles are similar. Thus we have shown all 4 triangles are similar to each other.

In each triangle, the longer leg to the shorter leg has a ratio of 4/3 (and the shorter leg to longer leg is a ratio of 3/4).

Suppose the square has a side length equal to s. Then DE = s. In triangle FEB, FE = s, so EB = (4/3)s. In triangle ADG, DG = s, so AD = (3/4)s.

Thus we have:

AB = AD + DE + EB
AC = (3/4)s + s + (4/3)s
5 = s(3/4 + 1 + 4/3)
s = 60/37
s² = 3600/1369 ≈ 2.63

Reference

This problem is a subset of 2017 AMC 10A, problem 21.
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_21

Previous post
https://mindyourdecisions.com/blog/2020/10/10/square-in-3-4-5-triangle/

Here is the setup:

Bob is an aspiring tennis player. He can earn a prize by winning (at least) 2 tennis sets in a row out of a three-set series. He has to play his teacher and the club champion alternately: either teacher-champion-teacher or champion-teacher-champion according to Bob’s choice. Everyone knows the club champion is a better player than the teacher. Which series should Bob choose?

As usual, watch the video for a solution.

Successive Wins Interview Question

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Answer To Successive Wins Interview Question

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

A game theory approach…

You can reach the answer even without doing any math. Just think like the problem maker and reason backwards.

Since the teacher is the worse player, it would make sense that Bob play him more than the champion. The choice of teacher-champion-teacher seems obvious… perhaps it’s a little bit too obvious.

This puzzle appears in a book of challenging probability problems, so the obvious answer should be held with suspicion. After ruling that choice, one can conclude the correct choice must be champion-teacher-champion.

(I know this sounds like that Princess Bride scene, but thankfully this reasoning has an ending. And in fact, champion-teacher-champion is the right answer.)

While we have reasoned to the correct answer, we have not solved a more interesting question: why should playing the tougher opponent more make Bob’s winning streak more likely?

The explanation

The reason has to do with the order of play. While champion-teacher-champion will have fewer expected wins, Bob doesn’t care about wins per se. He cares about creating a streak.

In this puzzle, Bob needs to win at least two sets in a row. How many ways is that possible? Begin by writing out the possible outcomes of the sets and identifying the win streaks. We’ll abbreviate for win (W) and loss (L):

WWW – prize
WWL – prize
WLW
LWW – prize
WLL
LWL
LLW
LLL

Notice that Bob has precisely three ways to make a streak: by winning all the sets (WWW), winning the first two (WWL), and winning the last two (LWW).

The key feature here is that Bob needs to win the second set in order to make a streak.

As the second set is critical, Bob would be wise to play the worse player–his teacher–in that slot. And that is the reason why champion-teacher-champion is the correct answer.

The size of the advantage

If we make a few more assumptions we can also calculate the size of the advantage.

Suppose Bob’s odds of winning a set depend purely on whom he plays, and suppose each set can be considered an independent event. Let t and c denote the probability Bob can beat his teacher and the champion in any given set, respectively.

Assuming independence of sets, we can then add up the probabilities of the winning events and compare their size:

Teacher-champion-teacher
WWW = tct
WWL = tc(1 – t)
LWW = (1 – t)ct
Total = ct(2 – t)

champion-teacher-champion
WWW = ctc
WWL = ct(1 – c)
LWW = (1 – c)tc
Total = ct(2 – c)

Since the club champion is a better player, c is less than t and 2 – c is larger than 2 – t. Thus ct(2 – c) is the larger quantity and it is better to play the champion twice.

Taking the example of t=0.8 and c=0.4, the chances are 51.2 percent for champion-teacher-champion and only 38.4 percent for teacher-champion-teacher. Choosing the right order increases the chances dramatically.

Special thanks this month to:

Kyle
Daniel Lewis
Lee Redden

Thanks to all supporters on Patreon and YouTube!

References

Old post
https://mindyourdecisions.com/blog/2008/12/09/a-math-problem-that-might-help-you-win-one-million-dollars/
Medium
https://medium.com/@rishidarkdevil/the-successive-wins-problem-f4d4397853d5
Fifty Challenging Problems in Probability with Solutions, By Frederick Mosteller
https://store.doverpublications.com/products/9780486134963?srsltid=AfmBOorI4gqlaBoO8CLEGhez4Wt5eqfmNi-CHlspPW6gZpEIauDA3YLN
Jane Street Capital interview question
https://www.glassdoor.com/Interview/suppose-you-player-A-player-B-are-in-a-tennis-tournament-Player-A-has-a-better-chance-of-beating-you-than-player-B-does-QTN_124563.htm