This puzzle has gotten popular because many people thought it was impossible to solve. Can you figure it out?

As usual, watch the video for a solution.

Inscribed Circle Puzzle

Or keep reading.

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Answer To Inscribed Circle Puzzle

(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 first step is to show the 4 right triangles are congruent. The angles of adjacent triangles are complementary, but so are angles in each of the right triangles, so all the acute angles in the right triangles are equal and the triangles are similar. As each hypotenuse is a side of the inscribed square, the hypotenuses are congruent. Similar triangles with a congruent side are congruent, so the 4 right triangles are congruent.

So all 4 triangles have an area of 24, and their sides can be labeled as a, b and c.

The inscribed’s square area is 100, so we have:

c² = 100
c = 10

In the large square, its area is equal to the square of its side length and the sum of the 5 shapes inside, so we have:

(a + b)² = 100 + 4(24)
(a + b)² = 196
a + b = 14

Finally we inscribe a circle in one of the triangles.

There are 2 formulas for the inradius we could use. For any triangle:

A = r(a + b + c)/2
24 = r(14 + 10)/2
r = 2

Or we can use that a right triangle’s inradius is given by:

c = a + b – 2r
10 = 14 – 2r
r = 2

In either case, the area of the inscribed circle is:

πr²
= π(2)²
= 4π

Deriving the inradius formulas

For any triangle:

For a right triangle:

Special thanks this month to:

Daniel Lewis
Lee Redden
Kyle

Thanks to all supporters on Patreon and YouTube!

Reference

Seen on Reddit AskMath
https://www.reddit.com/r/theydidthemath/comments/1r1yepq/can_you_solve_it_request/

As usual, watch the video for a solution.

How Tall Is The Man Puzzle

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Answer To How Tall Is The Man Puzzle

(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).

At first the puzzle seems impossible to solve as we do not know the distance between the two walls. Surprisingly, the distance does not matter! For poles of length a and b, the height of intersection h between the lines joining the top of one pole to the bottom of the other is equal to:

h = ab/(a + b)

We can set a = 4 and b = 6 to solve for the height of the man:

h = (4)(6)/(4 + 6) = 2.4 m

Proof of two poles formula

Label the two triangles are EFG and GJE, with the man replaced by the line segment KL. Let the distance between the two poles be d, and let KG = x so that KE = d – x, as shown below.

Triangles EFG and KLG are similar, giving:

x/h = d/a
x = dh/a

Triangles GJE and KLE are similar, giving:

(d – x)/h = d/b
bd – bx = dh
x = (bd – dh)/b

Both equations are equal to x, so setting them equal to each other gives:

dh/a = (bd – dh)/b
bdh = abd – adh
bdh + adh = abd
hd(b + a) = abd
h(b + a) = ab
h = ab/(a + b)

Special thanks this month to:

Kyle
Daniel Lewis
Lee Redden

Thanks to all supporters on Patreon and YouTube!

References

Puzzle
https://x.com/H0H0v/status/2007323697477210181

Two Poles Formula
https://artofproblemsolving.com/wiki/index.php/Two_poles_formula?srsltid=AfmBOootttF_6CfLUa7mr2vL_pqughPMQSXMEuLj0KEtGuWfJJ6E_JiR

Robert Wadlow
https://commons.wikimedia.org/wiki/File:Robert_Wadlow,_World%27s_Tallest_Man_Statue.jpg
KMOM14, CC BY 3.0 , via Wikimedia Commons

Becoming successful is not luck. It’s math.

If your probability of success is 1/100 and you try 100 times, you have a 100% chance of success.

For whatever reason many people decided this mistake had to be corrected. So one might ask: what is the actual chance of success? How many tries (rounded to the nearest 100) would give you a 99% chance of success?

As usual, watch the video for a solution.

Post About The Math Of Becoming Successful

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Answer To Post About The Math Of Becoming Successful

(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).

Let p = chance of success and 1 – p = failure. Then for n independent trials, the chance you have at least 1 success is:

Pr(at least 1 success in n trials)
= 1- Pr(all failures in n trials)
= 1- Pr(failure trial 1)Pr(failure trial 2)…Pr(failure trial n)
= 1 – (1 – p)ⁿ

For p = 1/100 and n = 10 we have:

= 1 – (1 – 1/100)¹⁰⁰
= ≈ 63.4%

In spite of 100 trials, you still have about a 36.6% chance of failure. Interestingly, when n = 1/p, we have:

1 – (1 – p)ⁿ
= 1 – (1 – 1/n)ⁿ
→ 1 – 1/e
≈ 63.2%

So interestingly as n goes to infinity we have a limit of success.

But increasing the number of trials does increase your chance of success. To get a 99 percent chance of success:

0.99 = 1 – (1 – 1/100)ⁿ
0.99ⁿ = 0.01
n = ln(0.01)/ln(0.99
n ≈ 458.2

Rounding up to the nearest 100, we would need 500 trials to get a 99 percent chance of success.

Leila Hormozi’s initial tweet was met with plenty of sarcasm and some name calling, even earning a “community note” to clarify the actual math. But she took the higher road and replied with the following tweet.

Community Note is right.
It’s ~63%, not 100%.

Somehow I’ve managed to function and become successful in business despite being atrociously bad at math. lol. not a secret you can ask my team.

Here’s what I meant to say:
1 attempt = 1% odds.
100 attempts = 63% odds.
500 attempts = 99.3% odds.

Persistence doesn’t guarantee success. It does compounds your probability until the math is eventually on your side.

And now we know that the worse you are at math…. the less time it takes🫣😂😅

Seeing a 99% failure rate, many people give up right away, and as the famous saying goes, you miss 100 percent of the shots you don’t take.

Some people will try for 100 times, but be disappointed by the 36.6% chance of failure.

But very few people will persist for 500 attempts and reap the high 99% chance of success.

The math of success is that failure is acceptable; but not trying is not acceptable. While luck will play a role, perhaps it is those ignorant of the odds who are happy to keep trying and will succeed against all odds.

References

Leila Hormozi
https://x.com/LeilaHormozi/status/2045498332928209202
https://x.com/LeilaHormozi/status/2046202486525211089

Zocchihedron Man, CC BY-SA 3.0
https://commons.wikimedia.org/wiki/File:Zocchihedron2.jpg

Math StackExchange
https://math.stackexchange.com/questions/165993/average-number-of-times-it-takes-for-something-to-happen-given-a-chance

Rober Mistake similar mistake years ago
https://x.com/MarkRober/status/1168950821570195456
http://mindyourdecisions.com/blog/2019/09/23/puzzle-inspired-by-mark-rober-tweet/