Loading...

Follow Mind Your Decisions on Feedspot

Continue with Google
Continue with Facebook
or

Valid

Recently I posted about a problem that was viral in China. Solve for the area shown in red, in the diagram below, between the semicircle and the rectangle’s diagonal.

Today’s challenge is to solve for the general case. Find a formula for the red shape in terms of r, the radius of the semicircle.

You can use the method I presented in the previous problem. But I will warn the algebra gets a little messy! It turns out there is a simpler way to solve the problem, and I credit Tony Hasler for the solution.

Watch the video to see the answer.

Viral Math Problem – Simple Formula For The General Solution

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Viral Math Problem – Simple Formula For The General Solution

(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).

Thanks to Tony Hasler for deriving this solution!

We start by setting up a coordinate system, where (0,0) is the lower left hand corner. The diagonal has the equation y = x/2 and the semicircle has the equation y = r – √(r2 – (xr)2).

Now let’s focus on where the semicircle and line y = x/2 intersect. We need to solve the following:

r – √(r2 – (xr)2) = x/2

r2 – (xr)2 = (x/2 – r)2
5x2 – 12xr + 4r2 = 0
(5x – 2r)(x – 2r) = 0
x = 2r or x = 2r/5

The solution x = 2r corresponds to the upper right hand corner of the rectangle. We want the interior intersection, which will be x = 2r/5.

Since y = x/2, the y-coordinate will be r/5. So the curves intersect at (2r/5, r/5).

Now draw a line connecting the points (r, r) and (2r/5, r/5). We can then write an equation for the area of the red shape as the sum of the areas of two triangles minus the area of a circular sector.

We just need to know where the line between (r, r) and (2r/5, r/5) intersects the x-axis. As this line has a slope of 4/3, it is straight-forward to calculate its x-intercept is (r/4, 0).

The blue triangle then has an area of:

blue triangle area
0.5(r/4)(r/5) = r2/40

The green triangle, with corners (r/4, 0), (r, 0), and (r, r) has legs of sides 3r/4 and r, so its area is:

green triangle area
0.5(3r/4)(r) = 3r2/8

Now all that remains is the area of the purple circular sector. We can find its central angle using the dimensions of the green triangle. Since the opposite side is 3r/4 and its adjacent side is r, the tangent of the angle is equal to 3/4. So the angle’s measure is arctan(3/4). A circular sector’s area is 0.5r2(central angle), so its area is:

circular sector area
0.5r2 arctan(3/4)

Now remember the calculation is:

We now add the areas of the blue and green triangles and subtract the area of the circular sector to get:

r2/40 + 3r2/8 – 0.5r2 arctan(3/4)

= r2(0.4 – 0.5 arctan(3/4))

And that’s it! We can solve for the area of this kind of share for any value of the radius, whether we want r = 4, or r = 10, or r = 73 (and of course, you know 73 is the best number).

It’s great when a hard geometry problem works out to such a simple formula.

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

Congratulations to Daniel Mai, 13, for winning the 2019 Raytheon MATHCOUNTS National Competition! The prize includes a $20,000 college scholarship.

In the final round of competition, Mai faced Suyash Pandit in a speed test. Only the first person to solve a problem got the point, and the first to 4 points wins. And the round is pencil and paper only, no calculators are allowed.

This is the question that sealed his victory.

What is the quotient of 5040 divided by the product of its unique prime factors?

Watch the video for a solution.

Teen Won $20,000 Solving This Question

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Teen Won $20,000 Scholarship Solving This Question

Let’s work out the prime factorization of 5040.

5040 = 2 × 2520
5040 = 2 × 2 × 1260
5040 = 2 × 2 × 2 × 630
5040 = 2 × 2 × 2 × 2 × 315
5040 = 2 × 2 × 2 × 2 × 3 × 105
5040 = 2 × 2 × 2 × 2 × 3 × 3 × 35
5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7

The unique prime factors are 2, 3, 5, 7, so we want to divide 5040 by the product of them, which gives:

5040/(2 × 3 × 5 × 7) = 2 × 2 × 2 × 3 = 8 × 3 = 24

The answer is 24.

But there’s another faster way! As commenters on YouTube pointed out to me,

5040 = 7! = 2 × 3 × 4 × 5 × 6 × 7

After removing the unique prime factors, we are left with 4 × 6 = 24. This is the better way to solve this in a speed competition!

Mai solved the problem in a matter of seconds–perhaps with the factorial trick–which is a credit to hard work and preparation. And who knows, judging by the intelligence of people reading MindYourDecisions, maybe next year’s MATHCOUNTS winner will be someone who’s reading this post right now 😉

Sources
Raytheon 2019 MATHCOUNTS press release
https://www.raytheon.com/news/feature/best-when-it-counts

ESPN webcast of 2019 MATHCOUNTS National Competition
http://www.espn.com/watch/_/id/3508357/2019-raytheon-mathcounts-national-competition?linkId=67110954

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

Thanks to Ata from Turkey for suggesting this! The problem is intended for the 8th grade (around 13 years old) and is adapted from a book of practice problems, Tonguc 8. Sinif Sayisal Zoru Bankasi).

Here’s the problem.

Tongul draws 2 rectangles ABCD and ABEF whose edges are integers (in cm).

The area of ABCD is 20 cm2 and ABEF is 10 cm2.

Which is the minimum possible perimeter of the rectangle CDFE in cm?

(a) 12
(b) 14
(c) 22
(d) 26

I admit I got this wrong! Can you figure it out? Watch the video for a solution.

Problem For 13 Year Olds That Stumps Adults!

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Math Problem For 13 Year Olds That Stumped Me

(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).

I’ll first present who I approached the problem and got to the wrong answer. I drew the two rectangles as follows.

If we set AB = x, then the opposite sides EF and CD are also equal to x. Since ABCD has an area of 20, BC = AD = 20/x. Since ABEF has an area of 10, BC = AD = 10/x.

The perimeter of CDEF is then:

x + 20/x + 10/x + x + 10/x + 20/x
= 2x + 60/x

Since each side is an integer, 10/x has to be an integer. Hence we can consider factors of 10 for values of x, so x could be 1, 2, 5 or 10. The perimeter values could be:

2(1) + 60/1 = 62
2(2) + 60/2 = 34
2(5) + 60/5 = 22
2(10) + 60/10 = 26

From these calculations, I thought the answer was 22. But this is not correct. Where is the mistake?

The correct answer

The trick is to think inside the box! The original diagram can be drawn with ABEF contained inside of ABCD like this:

We can now proceed similarly. Again letting AB = x, we have EF = CD = x. Then since ABCD has an area of 20, BC = 20/x = AD. Since ABEF has an area of 10, BE = AF = 10/x. We then must have EC = FD = BCBE = 10/x.

The perimeter of CDEF is then:

x + 10/x + x + 10/x
= 2x + 20/x

Again the values of x are the factors of 10, so we have:

2(1) + 20/1 = 22
2(2) + 20/2 = 14
2(5) + 20/5 = 14
2(10) + 20/10 = 22

Thus the minimum perimeter is (b) 14.

While I didn’t solve the problem, it was a fun one. We’re always told to think outside the box. But in a problem like this, you have to think inside the box!

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

A rectangle contains 3 identical circles as shown in the diagram.

(the top two circles are tangent to two sides of the rectangle; the bottom circle is tangent to one side of the rectangle; each circle is tangent to the other two circles)

If each circle has a radius of r, what is the area of the rectangle in terms of r?

The problem is adapted from one of the hardest GCSE exam problems (a test given in the UK).

Watch the video for a solution.

Rectangle Area From 3 Identical Circles – Solving Hard GCSE Problem

Or keep reading.

.
.
.
.
“Weave a circle round him thrice”
Samuel Taylor Coleridge, Kubla Khan
.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
Answer To Rectangle Area From 3 Identical Circles

(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 can solve for the side lengths by drawing radii of the circles.

The horizontal distance is equal to 4 radius lengths, or 4r.

The tricky part is the vertical distance between the centers of the circles. If we connect the centers of the 3 circles, we get an equilateral triangle with a side length of 2r. So its altitude is r√3. We then add 2 radius lengths to get the entire side length, which is then r(2 + √3).

Thus the area of the rectangle is:

(4r)r(2 + √3)
= 4r2(2 + √3)

Sources
Videos on YouTube, problem from the 2017 GCSE exam
https://www.youtube.com/watch?v=LnLALgbPbxQ
https://www.youtube.com/watch?v=ZN9Ty19Rzr8

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

Construct a hexagon from two congruent parallelograms with the following dimensions:

The problem is to solve for the cosine of PBQ in terms of x.

This comes from the 2017 GCSE exam, and it confused many people. I received many requests to solve this problem, and I thank Tom, Ben, and James for suggesting it to me.

Watch the video for a solution.

One Of The Hardest GCSE Test Questions – How To Solve It

Or keep reading.

.
.
.
.
“It’s more fun to arrive a conclusion than to justify it.” Malcolm Forbes
.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
Answer To One Of The Hardest GCSE Test Questions

(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).

Recall in any triangle c2 = a2 + b2 – 2ab cos C

This is often called the “law of cosines.” But in my research, I found a fun fact on Wikipedia:

In France, the law of cosines is named Théorème d’Al-Kashi (Theorem of Al-Kashi), as al-Kashi was the first to provide an explicit statement of the law of cosines in a form suitable for triangulation.

I think it makes sense to call the identity Al-Kashi’s Theorem, and I will try to call it that henceforth.

The solution

The problem is a test of applying Al-Kashi’s Theorem two times. First draw PQ and notice it has the same length if we shift the line segment to connect the corners of two parallelograms, as shown here:

We can use Al-Kashi’s Theorem in the upper triangle, noting the side opposite angle B has the same length as PQ. This gives:

PQ2 = x2 + x2 – 2(x)(x) cos 30°
PQ2 = 2x2 – 2x2(0.5√3)
PQ2 = x2(2 – √3)

Now let’s focus on triangle PBQ, recalling that we are given BP = BQ = 10.

We can use Al-Kashi’s Theorem to get:

PQ2 = 102 + 102 – 2(10)(10)cos PBQ
PQ2 = 200 – 200 cos PBQ

We now have two expressions equal to PQ2, so we can set them equal to each other. We get:

x2(2 – √3) = 200 – 200 cos PBQ

Subtract 200 from both sides and then divide by -200 and we get the answer:

cos PBQ = 1 – x2(2 – √3)/200

It’s a tricky problem since you have to use the law of cosines twice, and you have to keep track of a lot of algebra. I don’t think I’d solve this in a timed test. But it is a fun problem nevertheless!

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

I’ve adapted this problem from a GCSE problem that many students find very challenging.

A bag contains only red, gold, and green counters.

Alice takes at random a counter from the bag, and then she puts the counter back in the bag.

Bob does the same: he takes at random a counter and then returns it to the bag.

The probability that both counters are red or that both counters are gold is 5/16.

The probability that the first counter is red and the second counter is not red is 1/4.

Charlie takes at random a counter from the bag.

What is the probability that Charlie takes a gold counter?

Watch the video for a solution.

How To Solve A Tricky GCSE Probability Problem


.
.
.
.
“Every day is like survival” Karma Chameleon
.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
Answer To Tricky GCSE Probability Problem

(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 GCSE exam always has great probability questions like this one.

A problem like this can be intimidating because it contains a lot of details. One important step is reducing the problem to its mathematical concepts.

Since Alice returns the counter to the bag, and so does Bob, the problem is about sampling with replacement. Thus, the probability of drawing a particular color is the same–it doesn’t matter what the previous person took.

So let’s set up probabilities for drawing each color:

x = P(red)
y = P(gold)
z = P(green).

(Actually we don’t really need z since the probability only deals with red and gold–but you won’t know that until you work the problem out!)

Now let’s translate: “The probability that both counters are red or that both counters are gold is 5/16.”

The probability both are red is:

x(x) = x2

The probability both are gold is:

y(y) = y2

We sum these to get 5/16:

x2 + y2 = 5/16

Next let’s translate: “The probability that the first counter is red and the second counter is not red is 1/4.”

x(1 – x) = 1/4

This second equation involves only one variable, x, so we can solve for it.

x(1 – x) = 1/4
xx2 = 1/4
x2 – x + 1/4 = 0
(x – 1/2)2 = 0
x = 1/2

We then go to the other equation to solve y.

x2 + y2 = 5/16
(1/2)2 + y2 = 5/16
y2 = 1/16
y = 1/4 or y = -1/4 (reject)

The probability of drawing a gold counter is positive, so we take the positive root to the equation y = 1/4.

The probability Charlie draws a gold counter is thus 1/4.

Sources
Adapted from Twitter/Math StackExchange
https://twitter.com/MjEffGerAsoN/status/1122173214854135808
https://math.stackexchange.com/questions/3103891/gcse-probability-question-relating-to-trees-help-please

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

I thank a viewer from Edmonton, Canada for suggesting this problem. It is from a junior math competition from around 1970. Solve for the radius of the circle given the two perpendicular chords with lengths as shown in the diagram.

Can you figure it out? Give it a try, and then watch the video for a solution.

How to Solve For The Radius – Fun 1970s Math Contest Problem

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Solve For The Radius – Fun 1970s Math Contest Problem

(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).

I will present two ways to solve this problem: classic geometry and coordinate geometry.

Method 1: classic geometry

There’s an incredibly simple way to solve this problem. I learned about it in a video on the channel Maths By Amiya: Perpendicular Chords and Radius.

Consider the two perpendicular chords being divided into lengths w, x, y, z as shown:

Then the radius r is given by the formula:

4r2 = w2 + x2 + y2 + z2

In our problem, we would have w = 2, x = 6, z = 3, and we would need to solve for y.

We can solve for y by the power of the point/intersecting chords theorem, which states the product of the two segments are equal, so:

wx = yz
2(6) = y(3)
4 = y

So now we substitute:

4r2 = w2 + x2 + y2 + z2
4r2 = 22 + 62 + 42 + 32
4r2 = 65
r = 0.5√65 ≈ 4.031

This is great. But what if you didn’t know this nice formula? Let’s prove why it is true.

Proof of formula

The key is the perpendicular bisector of a chord always passes through the center of a circle.

(Why? The perpendicular bisector of the chord is the locus of points equidistant from the endpoints of the chord. The locus of such points has to include the circle’s center, as it is a fixed distance–a radius length–from the endpoints of the chord.)

So let’s draw the perpendicular bisector of each chord. Since each line contains the circle’s center, their single point of intersection has to be the circle’s center. Each perpendicular bisector passes through the midpoint of its respective chord, so we can also calculate the length to the midpoints for each chord as (w + x)/2 and (y + z)/2, and then solve for the other distances in the diagram:

Now we will draw a right triangle from the circle’s center where two of its legs are based on the above construction.

The hypotenuse goes from the circle’s center to the circumference, so it must be a radius of the circle, r. We then use the Pythagorean Theorem to get:

r2 = (0.5(wx))2 + (0.5(y + z))2
r2 = (w2 + x2 + y2 + z2)/4 – 0.5(wxyz)

Now recall wx = yz by the intersecting chords theorem, so the term wxyz vanishes. So we get:

r2 = (w2 + x2 + y2 + z2)/4

Multiplying both sides by 4 gives the formula presented before.

Method 2: coordinate geometry

I did not know the neat formula, so I solved it using coordinate geometry. I set up a coordinate system where the origin (0, 0) is the intersection point of the two chords, and the circle has a center (p, q):

Using the general equation for a circle, from point A we get the equation:

(-2 – p)2 + q2 = r2

Then for point B we get the equation:

(6 – p)2 + q2 = r2

Subtracting the second from the first, we eliminate the variables q and r, so we get:

(-2 – p)2 = (6 – p)2
p = 2

Substituting the value for p gives:

42 + q2 = r2

We now find an equation for point D, using that p = 2 to get:

22 + (-3 – q)2 = r2

Subtracting the equation for D from B gives:

12 + q2 = (-3 – q)2
6q – 3 = 0
q = 0.5

We substitute that into any of the point equations gives:

16.25 = r2
r = 0.5√65 ≈ 4.031

I find both ways to solve the problem to be interesting, and it’s a pretty fun problem overall!

Thanks to all patrons! Special thanks to:

Michael Anvari
Richard Ohnemus
Shrihari Puranik
Kyle

I also credit patrons Pradeep Sekar and Nestor Abad for finding a typo in my original video–thanks! (You can get early access to videos by supporting on Patreon–such support makes a huge difference.)

Help inspire mathematical discovery around the world! Support with a pledge on Patreon: http://www.patreon.com/mindyourdecisions

Reference

Maths By Amiya : Perpendicular Chords and Radius
https://www.youtube.com/watch?v=cSkswi-Y0x4

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

Mathematics isn’t usually part of popular culture. But in 1982 the mathematician Raymond Smullyan appeared on Johnny Carson’s Tonight Show and shared a couple of logic puzzles.

Puzzle 1

John, James and William are triplets who look the same. One day you meet one of them on the street, and you want to figure out if the person is John.

You know that John and James always lie, and William always tells the truth. You can ask a single question, whose answer has to be yes or no, and the question can have at most 3 words. What could you ask?

Puzzle 2

There are one hundred politicians at a party. Each politician is either honest or crooked, and they are not all of the same type—-at least one is honest and at least one is crooked.

If you pick any two politicians at random, at least one in the pair is crooked. How many honest politicians are there?

Watch the video for a solution.

Logic Puzzles From The Tonight Show, 1982

Or keep reading.

.
.
.
.
“I continue to write puzzle books, but these are more than mere puzzle books — it is through recreational logic puzzles that I introduce the general reader to deep results in mathematics and logic!” Raymond M. Smullyan
.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
Answer To Logic Puzzles From The Tonight Show

(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).

It is pretty fun to watch how Johnny Carson tried to solve the puzzles, and Raymond Smullyan was directing him towards the correct answers. So do check that out: Raymond Smullyan appeared on Johnny Carson’s Tonight Show

Puzzle 1

Johnny Carson attempted the problem and explained his thought process. First he considered the question:

“Does John lie?” But this question doesn’t work since both John and James answer “no”:

“Does John lie?”
John (lies) – No
James (lies) – No
William (tells truth) – Yes

Then Carson considered:

“Are you John?” This is closer, but still doesn’t work since John and William answer “no”.

“Are you John?”
John (lies)- No
James (lies) – Yes
William (tells truth) – No

Smullyan suggests Carson to modify the question just a little bit so that John has a unique answer. Then Carson figures out the answer:

“Are you James?”*
John (lies) – Yes
James (lies) – No
William (tells truth) – No

(*answers corrected thanks to email from Mike!)

So it’s a little bit of a funny situation. To figure out who John is, you have to ask about James and not John!

Puzzle 2

Carson thinks about this and figures out the answer relatively soon.

The answer is there is just 1 honest politician. If you had 2 (or more) honest politicians, then any 2 of them would be a pair of honest politicians, which contradicts that every pair has at least 1 crooked politician.

Thus there can be at most 1 honest politician. Since the puzzle states there is at least 1 honest politician, we can conclude there must be exactly 1 honest politician.

Could this happen today?

A few people might ask, “Would a popular late night talk today show ever feature a mathematician?”

The answer is a resounding YES! A couple of years ago Stephen Colbert featured Dr. Eugenia Cheng, author of How To Bake Pi.

So it’s not impossible for mathematicians to be in popular shows. But it is less common than I would like.

Sources
Raymond Smullyan on The Tonight Show Starring Johnny Carson
https://www.youtube.com/watch?v=E27v83WWiGo

The Lady or the Tiger?: and Other Logic Puzzles (book that Raymond Smullyan was promoting on the Tonight Show)
I may get a commission if you purchase from this link, but it doesn’t affect the price you pay:
https://amzn.to/2ZnxkEZ

Wikipedia
https://en.wikipedia.org/wiki/Raymond_Smullyan

Princeton obituary
https://paw.princeton.edu/article/lives-raymond-m-smullyan-59

h/t FiveThirtyEight Riddler
https://fivethirtyeight.com/features/a-riddle-how-many-crooked-politicians-are-there/

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 
Mind Your Decisions by Presh Talwalkar - 1M ago

A gigantic pie is split amongst 100 guests. The first guest gets 1% of the pie. Then the second guest gets 2% of the remaining pie. Then the third guest gets 3% of the remaining pie, and so on, until the last guest gets 100% of the remaining pie.

Which guest gets the largest piece of pie?

The challenge is to solve without using calculators or computers which would numerically calculate the answer very quickly.

(The connection to Pythagoras is this problem appeared in the Pythagoras Magazine, per an article in The Guardian.)

Watch the video for a solution.

Can You Solve The Pythagoras Pie Puzzle?

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To The Pythagoras Pie 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).

Let’s calculate the fraction of the entire pie that each guest gets.

Guest 1 gets 1/100 of the pie. Guest 2 is left with 99/100 of the pie and gets 2/100. Guest 3 is left with (99/100)(98/100) and gets 3/100. So we see a pattern, and we can write a general formula for guest k:

Gk = (99/100)(99/100)…((99 – (k – 2))/100)(k/100)

Gk = (99!/(100 – k)!)(k/100k)

We want to find where this function is increasing, so we want the ratio between two consecutive guests.

Gk+1/Gk
= (99!/(100 – (k + 1))!)((k + 1)/100k+1)/[(99!/(100 – k)!)(k/100k)]
= (100 – k)(k + 1)/(100k)

The pie slice is increasing if and only if this fraction is larger than 1, which means the condition:

1 k)(k + 1)/(100k)
k2 + k k = 1, 2, …, 9. So at k = 9, we have G10/G9 > 1, so the largest slice of pie goes to guest 10. (After that the pie slice will decrease in size.)

So the answer is guest 10 gets the largest piece of pie.

As a fun curiosity, we can calculate:

G10 = (99!/90!)(10/10010) = 62815650955529472/1009 ≈ 6.28%

As a fun fact, this puzzle about pie has an answer that is approximately 2π/100. It’s just a coincidence, but it’s a nice way to remember the answer to this fun puzzle!

Sources
The Guardian (puzzles from Pythagoras Magazine)
https://www.theguardian.com/science/2017/jun/19/did-you-solve-it-pythagorass-best-puzzles

Puzzling StackExchange
https://puzzling.stackexchange.com/questions/52787/who-get-the-biggest-piece

Read Full Article
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

Here’s a fun problem I saw on Twitter from Peter Rowlett. Solve this equation. Then turn the equation upside down and solve again!

Watch the video for a solution, and to learn more about ambigrams.

The Incredible Upside Down Equation – Math Ambigram

Or keep reading.

.
.
.
.
A great truth is a truth whose opposite is also a great truth. Thomas Mann
.
.
.
.

M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
Answer To The Incredible Upside Down Equation

(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).

In 1908, the British magazine Strand described an interesting curiosity: the word chump reads the same upside down as it does right side up!

Chump is an example of an ambigram:

An ambigram is a word, art form or other symbolic representation whose elements retain meaning when viewed or interpreted from a different direction, perspective, or orientation. [Wikipedia ambigram]

Ambigrams are not limited to words. Here’s a mathematical ambigram, found by A.P. Goucher. The equation is not only true, but it is the same when you turn it upside down!

My personal favorite math ambigram is one that I discovered. If you use the multiply by lines method, you create a square lattice, which has the symmetries of the square. So you can rotate and reflect the diagram to solve even more problems! Each column is a 90 degree rotation clockwise, and the second row starts with the diagram for 12×13 reflected vertically about its center.

Not many people know of this discovery yet! But I am sure this “Talwalkar square” will be a standard part of math classrooms once the right people learn about it. For details see multiply by lines group theory.

The original problem

The first equation is:

(x + 8)/6 = 9/(5 + x)

We’ll solve it by cross-multiplying and then factoring the resulting quadratic.

(x + 8)(5 + x) = 6(9)
x2 + 13x – 14 = 0
(x – 1)(x + 14) = 0
x = 1 or x = -14

After we rotate the equation, we would get the fraction:

(5 + x)/6 = 9/(8 + x)

We solve as before:

(5 + x)(8 + x) = 6(9)
x2 + 13x – 14 = 0
(x – 1)(x + 14) = 0
x = 1 or x = -14

Remarkably we have the same solution as before!

This is an interesting result, and it’s not that simple to find an equation. Read the long process of creating the upside down equation.

This is a delightful quadratic equation, and it opened our eyes to the wonderful world of ambigrams!

Sources

Peter Rowlett Twitter
https://twitter.com/peterrowlett/status/1074599768993787904

Wikipedia ambigram
https://en.wikipedia.org/wiki/Ambigram

A. P. Goucher ambigram
https://cp4space.wordpress.com/2013/01/17/assorted-stuff/
via
https://twitter.com/MathsEdIdeas/status/1013877452702191617

Multiply by lines group theory ambigram “Talwalkar square”
https://youtu.be/LmVt_KiyDC4

Finding the upside down equation
https://aperiodical.com/2018/12/finding-an-equation-that-has-the-same-solution-when-rotated/

Read Full Article

Read for later

Articles marked as Favorite are saved for later viewing.
close
  • Show original
  • .
  • Share
  • .
  • Favorite
  • .
  • Email
  • .
  • Add Tags 

Separate tags by commas
To access this feature, please upgrade your account.
Start your free month
Free Preview