What is the sum of the corner angles in a regular 5-sided star? What is *a* + *b* + *c* + *d* + *e* = ?

Here’s a bonus problem: if the star is not regular, what is *a* + *b* + *c* + *d* + *e* = ?

Can you figure it out? Watch the video for a solution.

**What Is The Sum Of Angles In A Star? Challenge From India**

Or keep reading.

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**Answer To Sum Of Angles In A Star**

(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 the video, I explain an intuitive way to see the answer is 180 degrees. If you place a pen along one of the sides, and then rotate it through the 5 angles, you will end up with the pen in the same spot but flipped 180 degrees. You can see an animation of that here: star pentagon angle sum animation.

While that is not a proof, it suggests an answer of 180 degrees, which might give you the idea to think about half of a circle or the sum of angles in a triangle. These concepts can be used to prove the result.

**For a regular star pentagon**

There is a wonderful proof for a regular star pentagon. A regular star pentagon is symmetric about its center so it can be inscribed in a circle. From there, we use the fact that an inscribed angle has a measure that is half of the arc it subtends. So we get a figure like this:

Since a circle measures 360 degrees, and the arcs combine to be the entire circle, we get:

2*a* + 2*b* + 2*c* + 2*d* + 2*e* = 360 degrees

Now we divide this equation by 2 and magically we have the answer!

*a* + *b* + *c* + *d* + *e* = 180 degrees

Now you could extend this proof for the more general case of an irregular star pentagon. You could do this by showing the sum of the corner angles is unchanged as you move a corner (proof here). Thus you can always re-arrange an irregular star pentagon into a regular one, and since the total angle sum is unchanged, the irregular one must also have a measure of 180 degrees.

But there’s another proof that is more direct that I prefer.

**Proof for any star pentagon**

The key is to consider triangles. The exterior angle to the triangle with corners *b* and *d* has an angle measure *b* + *d*:

Similarly, the exterior angle to the triangle with corners *c* and *e* has an angle measure *c* + *e*:

Thus the top triangle has angles *a*, *b* + *d*, and *c* + *e*:

Since the sum of the angles in a triangle is 180 degrees, and this triangle has the sum of all the corner angles, we are done!

*a* + *b* + *c* + *d* + *e* = 180 degrees

There are many other ways to prove the result too! And you can then investigate other star polygons and closed curves–see the “further reading” link.

**Thanks to all patrons! Special thanks to:**

Shrihari Puranik

Richard Ohnemus

Michael Anvari

Kyle

Let’s math the world a better place! You can support these videos and posts at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Websites consulted while making video**

https://puzzling.stackexchange.com/questions/17681/five-angles-in-a-star

https://www.futilitycloset.com/2015/04/25/star-power-2/

http://mathandmultimedia.com/2012/06/23/angle-sum-of-pentagram/

http://mrhonner.com/archives/14821

http://mathandmultimedia.com/2012/06/23/angle-sum-of-pentagram/

http://proofsfromthebook.com/2013/08/04/angle-sum-of-a-pentagram/

https://en.wikipedia.org/wiki/Star_polygon

Further reading

http://www.math.nsysu.edu.tw/~wong/papers/soa-SEAM-formatted.pdf

A number lock requires a 3 digit code. Based on these hints, can you crack the code?

682 – one number is correct and in the correct position

645 – one number is correct but in the wrong position

206 – two numbers are correct but in the wrong positions

738 – nothing is correct

780 – one number is correct but in the wrong position

Watch the video for a solution.

**Can You Crack The Code? “Only Geniuses Can Solve”**

Or keep reading.

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**Answer To Can You Crack The Code? “Only Geniuses Can Solve”**

(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).

There are many ways to get to the code of 052. Here is one explanation.

738 – nothing is correct

We can eliminate the numbers 7, 3, and 8.

682 – one number is correct and in the correct position

Since 8 is incorrect, we know 6 or 2 is a correct number and correctly placed (but not both correct).

645 – one number is correct but in the wrong position

If 6 was correct and correctly place, then the code 645 should also have at least one number correctly placed. Thus we can conclude 2 was correctly positioned and correct in the code 682, and that 6 is an incorrect number. So we also know either 4 or 5 is a correct number, but wrongly positioned.

206 – two numbers are correct but in the wrong positions

Since 6 is wrong, the numbers 2 and 0 must be correct. From 682 we know 2 is correctly positioned in last spot. As 0 is not correctly positioned in the middle, it must be correctly positioned in the first spot. The code must be something like 0x2.

645 – one number is correct but in the wrong position

We know the code is 0x2. If the middle number was 4, then 645 would have one correct number in the correct position. Since it does not, and we know 6 is incorrect, it must be that 5 is the correct number.

Thus the number is 052.

**Sources**

https://www.reddit.com/r/puzzles/comments/9aelln/combination_lock_riddle/

https://puzzling.stackexchange.com/questions/46871/crack-the-lock-code

]]>The problem is to solve for the radius *r* in terms of *a*, *b*, and *c* in the following diagram.

Can you figure it out? Watch the video for a solution.

**Solve For The Radius – Harder Than It Looks!**

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**Answer To Solve For The Radius – Harder Than It Looks!**

(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).

First we make some chords in this circle.

Chord 1: extend the line segment with *c* to be a chord of the circle. The portion parallel to *a* has length *a*, and by symmetry the remaining line segment has length *c*.

Chord 2: extend the line segment with *b* to the circle’s circumference. We don’t know this line segment’s length, so we will label it *d*.

By the power of a point (aka intersecting chords theorem), we have:

*bd* = *c*(*a* + *c*)

We can solve for *d*:

*d* = *c*(*a* + *c*)/*b*

Next draw a line segment opposite the right angle between *a* and *b*, as shown below. Since this chord is opposite an inscribed right angle, this chord must be a diameter of the circle with length 2*r*.

We can then use the Pythagorean Theorem to conclude:

*a*^{2} + (*b* + *d*)^{2} = (2*r*)^{2}

We can substitute in for *d* to get:

*a*^{2} + (*b* + *c*(*a* + *c*)/*b*)^{2} = (2*r*)^{2}

Now it is a matter of simplifying this expression and then solving for *r*.

*a*^{2}*b*^{2}/*b*^{2} + ((*b*^{2} + *c*(*a* + *c*))/*b*)^{2} = 4*r*^{2}

(*a*^{2}*b*^{2} + (*b*^{2} + *c*(*a* + *c*))^{2}/*b*^{2} = 4*r*^{2}

(*a*^{2}*b*^{2} + *b*^{4} + 2*b*^{2}*c*(*a* + *c*)+*c*^{2}(*a* + *c*)^{2})/*b*^{2} = 4*r*^{2}

(*a*^{2}*b*^{2} + *b*^{4} + 2*b*^{2}*c**a* + *b*^{2}*c*^{2}+*b*^{2}*c*^{2}+*a*^{2}*c*^{2} + 2*ac*^{3} + *c*^{4})/*b*^{2} = 4*r*^{2}

The first four terms in the numerator have a common factor *b*^{2}, and the next four terms have a common factor *c*^{2}. So we have:

*b*^{2}(*a*^{2} + *b*^{2} + 2*ac* + *c*^{2})

*c*^{2}(*b*^{2} + *a*^{2} + 2*ac* + *c*^{2})

Both *b*^{2} and *c*^{2} are being multiplied by the same common factor, which can be factored as:

*b*^{2} + *a*^{2} + 2*ac* + *c*^{2} = *b*^{2} + (*a* + *c*)^{2}

So we get:

(*b*^{2} + *c*^{2})(*b*^{2} + (*a* + *c*)^{2})/*b*^{2} = 4*r*^{2}

Now we divide both sides by 4 and take the positive square root since *r* is positive, and we get:

(1/(2*b*))√((*b*^{2} + *c*^{2})(*b*^{2} + (*a* + *c*)^{2})) = *r*

It’s not the easiest formula to figure out, but this is the kind of intricate problem that comes out of math Olympiad problems.

**Sources**

QED Math blog Facebook page

https://www.facebook.com/qedmathblog/posts/1968955259865026

IMO 2016 Facebook page (sharing QED Math blog’s post)

https://www.facebook.com/imo2016/posts/2074333835931426

(hat) + (hat) x (pie) = 54

(pie) x (hat) + (pie) = 50

(hat) + (pie) + (hat) = ?

Is this too hard for a 7 year old?

I came across the problem via a tweet by Celeste Ng, who is the author of *Everything I Never Told You* and *Little Fires Everywhere*. Both books made the *New York Times* bestseller lists, and *Little Fires Everywhere* is being adapted into a TV.

Celeste Ng tweeted that this challenge was given to her 7 year old. And the problem caused arguments, even amongst people who could solve it, because it seemed too hard for 7 year olds.

The problem comes from Greg Tang, who is called the “math missionary” for his work in helping kids and adults with math. Since 2001 he’s done over 3,000 conferences and events, and accumulated over 1 million frequent flyer miles on 4 different airlines! Do check out GregTangMath.com to find math books, games, and other resources.

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

**Viral Problem – Solve Using Only Mental Math!**

Or keep reading.

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**Answer To Solve Using Only Mental Math!**

The standard way to solve the problem is using algebra. Let *h* = hat and *p* = pie. We then have the equations:

*h* + *hp* = 54

*ph* + *h* = 50

*h* + *p* + *h* = ?

First we note *hp* = *ph*, so the first two equations are:

*h* + *hp* = 54

*hp* + *p* = 50

We can subtract the second from the first to cancel the *hp* term and get:

*h* – *p* = 4

*h* = 4 + *p*

We can then substitute *h* = 4 + *p* into the second equation and simplify:

(4 + *p*)*p* + *p* = 50

*p*^{2} + 5*p* = 50

*p*^{2} + 5*p* – 50 = 0

Now we think of 2 factors of 50 whose difference is 5. So we can think of 5 and 10, so we can factor into:

(*p* + 10)(*p* – 5) = 0

Thus we get two solutions:

*p* = 5

*p* = -10

Since *h* = *p* + 4, we have:

*p* = 5, *h* = 9

*p* = -10, *h* = -6

So the final equation actually has two possible values.

If *p* = 5, *h* = 9

*h* + *p* + *h* = 23

(The problem was posted in 2017, and Thanksgiving Day in America was on Nov 23, 2017)

*p* = -10, *h* = -6

*h* + *p* + *h* = -22

(Perhaps since Nov 22, 2018 is Thanksgiving Day?)

**Can you solve with mental math?**

It’s a little bit trickier to solve using only mental calculations. But it’s possible! I do like to practice these skills once in a while as brain training.

The mental calculation starts the same way. Think about each picture as a variable, and translate each line into an equation in your head. Then you notice you can cancel *hp* by subtracting the second from the first. So in your head you get to the equations as before:

*h* = 4 + *p*

*p*^{2} + 5*p* = 50

But now how can you proceed? It’s not as easy to do the algebra steps in your head. So you might instead consider to guess and check for integer solutions. Re-write the equation as follows:

*p*(*p* + 5) = 50

We can consider factors of 50, and we want 2 factors that differ by 5. So we think about:

50 = 1 x 50 = 2 x 25 = 5 x 10

The solution *p* = 5 jumps out and indeed 5(5 + 5) = 50.

But what about the other solution? We can think about negative numbers! Notice that

50 = -1 x -50 = -2 x -25 = -5 x -10

Now the solution *p* = -10 jumps out and indeed -10(-10 + 5) = 50.

So now we use *h* = 4 + *p* to get the pairs of solutions *p* = 5, *h* = 9 and *p* = -10, *h* = -6.

We could then get *h* + *p* + *h* is either equal to:

23 (*p* = 5, *h* = 9)

or

-22 (*p* = -10, *h* = -6)

This is not a problem I’d find all that easy to solve in my head. But it’s also not impossible!

**Source**

Celeste Ng

https://twitter.com/pronounced_ing/status/935661930982838273

https://www.celesteng.com/about/

Greg Tang Math

https://gregtangmath.com/

https://twitter.com/gregtangmath

What is the value of AB in the following diagram?

Can you figure it out? Watch the video for a solution.

**Viral Facebook Puzzle. What Is AB = ?**

Or keep reading.

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**Answer To Viral Facebook Puzzle. What Is AB = ?**

The problem is equivalent to the horizontal distance between the centers of the blue and red circles. We can find the answer by adding the horizontal distance between the green and red circles, and then the blue and green circles.

First consider the green and red circles. Draw a right triangle whose legs are the horizontal and vertical distances between the centers, and whose hypotenuse connects the two circles.

One leg is the difference of radii, 3 – 2 = 1.

Because the circles are tangent, the hypotenuse is the sum of the radii, 3 + 2 = 5.

The other leg can be found by the Pythagorean Theorem, giving a value of √(5^{2} – 1^{2}) = 2√6.

This gives 2√6 as the horizontal distance between the green and red circles.

Now we find the horizontal distance between the blue and green circles. Draw a right triangle whose legs are the horizontal and vertical distances between the centers of those circles, and whose hypotenuse connects the centers of the two circles.

One leg is the entire rectangle’s vertical side minus the radii of the blue and green circles, 3 + 3 – 1.5 – 2 = 2.5.

As the blue and green circles are tangent, the hypotenuse is the sum of the radii, 1.5 + 2 = 3.5

The other leg can be found by the Pythagorean Theorem, which gives a value of √(3.5^{2} – 1.5^{2}) = √6.

This gives √6 as the horizontal distance between the blue and green circles.

Finally, the horizontal distance AB is the sum of the two horizontal distances we just calculated. Hence:

AB = √6 + 2√6 = 3√6 ≈ 7.35 cm

It’s great to see when geometry problems get popular!

**Reference**

If you know the original source please let me know. I was told this was popular on Facebook and Twitter, so I will provide a link to one example:

https://twitter.com/_WTProject/status/1040267815163441152