Advanced college geometry and we’re still calling sectors pizza slices. Never change math.

"AND THATS A GOOD PLACE TO STOP", the fourth brother said calmly.

Me after a long day of studying for finals: "let's get my daily dose of KZfaq to de-stress." KZfaq algorithm: here's some math completely unrelated to your degree Me: "hmmm yes, I'd like to know what that shaded area is too"

now shift the triangles altitude line 1% away from the center of the circle, rotate it clockwise 1 degrees and do it again

Me who's terrible at math: "Ah yes, of course! Triangles have THREE sides... It's so obvious now!"

I managed to follow this perfectly, right up until to the point he said "Hello there...".

As a wise man once said: "I wish I were high on potenuse"

The fact that the shaded area was found without the aid of integral calculus is a bit surprising.

Before seeing the video, I was planning to do it by calculus: 1. Set the bottom left point as the origin 2. Make equation for the circle: (y+r)^2+x^2=r^2 3. Make equation for the line: y=mx+b, where m=-(3+2.sqrt3)/(2+sqrt3) and b=2+sqrt3. 4. Find radius by differentiating the circle and setting it equal to to the slope, m. 5. Set the two equations equal to get the x component of the point of intersection (point of tangency) 6. Integrate the difference between the equation of the line and the equation of the circle over the range of 0 to the X value that we just found, the result should be the area.

The feeling of pride solving it and having the same answers but different solutions.

I think this solution complicated things a bit. When you find the first Pi/6, you will already know the center angle is Pi/6(similar triangle or simply calculate the shared pi/3 angle), and the edge is 1 by 4+2sqrt(3) - 3+2sqrt(3)

Hey Michael, I am from India. In my school days I was taught to do a perfect square factorization under a square root sign. It is the same method you used ( taking terms and equating to a and b) but it's done in shorter steps which may be a difficult for beginners or students who lack practice. What we do is this - 2ab√3 = 16√3 ab =8√3 a^2 + b^2 = 28 We got two equations. Now start by selecting values for a and b ( a =√3 , b = 8, a^2 + b^2 >28 a = 2√3 , b = 4, a^2 + b^2 = 28.... So a=2√3 and b=4

There is a much easier solution. The key is to see that the triangle formed by connecting the origin of the circle to the point of tangent is similar to the larger triangle. This allows you to solve for R. Also, since the ratio of the legs of the large triangle is SQRT(3) (after rationalizing the denominator), the triangle is a 30-60-90 triangle. You can then use R and the angle to obtain the area.

Math is beautiful I dont even speak english, but I understood everything

Once you found the third side, I quickly realized that the triangle was a 30-60-90 triangle. 30-60-90 triangles have the property such that the longer leg of the triangle is sqrt(3) times longer than the short leg, and the hypotenuse is 2x as long as the shorter leg.

Me: Who has no idea what he's doing Also me: Yep, seems about right.

The happiness when you figure it out correctly on your own and it turns out to be a simpler method.

This has to be first use of the half angle formula I’ve ever seen.

Great video! I was taken by surprise by your "Good place to stop", cause I thought you would make a common denominator for the solution.

That was some crazy math! So simple with Trig, algebra, definitions, and geometry, and yet so complicated! Thank you Professor Penn!

The most colorful Michael Penn video so far.

I had concluded my graduation in mechanical engineering 11 years ago, and Google still suggesting mathematics subjects at 1:00Pm. - Congratulations by clear explanation 👏🏽👏🏽👏🏽👏🏽👏🏽

and then in the corner of the paper, you see: *NOT DRAWN ACCURATELY*

It's 3 AM on a sunday. I haven't gone to school in 6 years orso. Tomorrow I have a D&D session I still need to prep for. And before this video I was watching Clips from the Castlevania netflix series. My brain: "Mmm indeed, How DOES one calculate that area?" KZfaq really is something else.

I'm physicist and this channel kept my attention with many interesting problems, thank you Michael, great stuff!

I love these videos. I love how I remember bits and pieces of algebra and trig from high school along the way and also learn new things I never was exposed to in school. Please keep it up!

this appeared randomly on my reccomended and it reminded me how much i love geometry and math, this was fun to understand

"okay so we are going to this this in a couple of steps" Ahh here we go again

This is the first video I’ve seen of you. And my mind exploded, my eyes widened! I love seeing how everything comes together. I am astonished, this makes me more interested in geometry. I loved this video, it made me really happy! :)

The main problem is calculate the “radius” And the “radius” could be calculated much more easily !! Each tangent leg /line to the circle is exactly of same length ( by definition) = 3+2√3 Since hypotenuse (h) = 4+2√3 Then, h - 3+2√3 = ( 4+2√3 ) - ( 3+2√3 ) = 1 *“That 1”* is the length ( of one of the leg) of the small “Top Triangle “ ( *which is congruent with the big triangle* ) The other Leg is “r “ ( radius) So 1/r = ( 2 + √3 ) / ( 3+2√3 ) Then r = ( 3+2√3 ) / ( 2 + √3 ) = √3 ========= *Note* If we see The small “Top Triangle “ is the half of an Equilateral Triangle with each side = 2 and “ height” = √3 So each angle is 60º Then the top angle of the small figure is 60ª ( 1/6 of the 360º) Then, everything is very EASY ! and you don't need trigonometry !! =========== Good moment to remember Einstein : *Imagination is more important than knowledge* *KISS principle* ¡!

This helped in revising my concepts of geometry & Trigonometry.

There is a simpler method on 6:21 , which is by simply splitting 28 such that it forms the addition of two squares i.e. 4 and 2 root 3. I would say it's simpler as it is a direct method and doesn't require many steps. Anyways, nice video.

To find the length 1 at 13:10 you could also use the theorem that tangent lines at a circle are equal. 4+2sqrt3-3-2sqrt3=1

Just a comment on style - I would’ve noted that the triangle is similar to the standard 1-2-sqrt3 special triangle, done all computations in terms of those simpler quantities, and then multiplied the answer by the factor of 2+sqrt3

I saw the thumbnail, and I wondered if we'd be paramaterizing the problem for a general solution amd started w/o watching. I've now watched for some insights bc I knew Michael knew much more what he was doing than I would know myself. The closer here is way better than what I started to do. I was going to compute a general scalene triangle formed by chords from the original right vertex to the tangent of the hypotenuse of the triangle and from said tangent to the diameter of the circle where the circumference crosses the leg it's symmetric about. Which was part one, then add the adjoined chord area at the end of this pseudo sector so ultimately I could subtract the sum from the encompassing triangle which includes the similar right triangle and true sector that I'll use now. Man, that just makes too much sense- thanks for that👍

Condense way: Step 1) You should know the two congruent triangle by tangent property Step 2) Get the angle (pi/12) and the hence the radius of circle by trigonometry Step 3) Final Area = Big triangle - Area of 2 congruent triangles - Area of half circle + 2 * Sector area with angle(pi/2 - pi/12)

tan(θ) = (2+√3)/(3+2√3) = √3/3 -> θ = pi/6. That saves 5 minutes.😃

General solution for triangle with sides a,b,c; side a being the side going through the circle: T = ((r^2)/2)((a/b)-arctan(a/b))

Thank you very much for demonstrating.

Excellent!! Like the pace of the proof.

The first several minutes - I immediately noticed that the long leg is just root3 times the short leg. That made this a lot easier. Also after 10:00 - the circle has two tangents meeting at the pi/6 vertex. They must be collinear. Thus as one is 3 + 2root2, the other must be, so there's an excess on the left of the point of tangency of 1. That's when this got really easy.

I never thought of placing a nested root equal to a + b√n and solving for a and b. Factoring an integer (like -4) and solving for each factor as a system of equations was also a new idea. Quite interesting techniques.

Loved it. More please. Thanks.

These videos are so entertaining when you are procrastinating. Also this man have a good voice and detailed instructions, makes people feel calm.

Is NO ONE going to comment on the brilliant Harry Potter reference made in the title?! I guess I have to.

To find the radius r and the angle alpha, use the similarity between the left-top triangle and the big green triangle (AA) would be simpler. :)

My reaction seeing the problem: "I see no obvious trick.... I am going to parametrize the **** out of this thing !" A brutal solution. But hey, it works !

That was so fascinating to watch. I’m in awe

13:51 You could find alpha by noticing that alpha plus the other two angles is a straight line and you already know the other two angles are both pi/4 - pi/12 = pi/6. So alpha = pi/2 - pi/3 = pi/6.

Me: Thinks I'm okay at math and want to proceed with something surrounding math in the future. Michael Penn: Crushes my dreams

i enjoyed the problem and your explanation ! great job professor .

That was beautiful! Thanks!

I have to admit, my math education only went up to Calculus 4 and differential equations in college, and after going into medicine none of this makes sense anymore. Definitely a good career move!

5 am, and for some random reason i am watching this...looked so hard in the beggining, thx for making it look easy at the end

A lot of your content I know I could do in high school, or would be able to figure out today if I had some more time. I've done that with a few of your videos that looked interesting or fun to solve at the start. This is the first video that not only would I not have been able to solve this in high school, I'm not even sure I could have solved it with some of my old college knowledge. Always fun to learn how to break down a complicated problem though!

13:50 : one can find triangle area by heron formula and alpha is trivially equal to theta, pi/6 so the arc area can be computed.

generalized formula to find the area (with a being the vertical length of the triangle and b being the horizontal length) ab/2-(b(b(tan(tan^-1(a/b)))-(((tan^-1(a/b))/360)(b(tan(tan^-1(a/b))2))

9:47 There is a more easy way to find the radius. The small triangle at the top is also rectangular and the top corner is pi/3. tan(pi/3) = r / (4 + 2 * sqrt(3) - 3 - 2 * sqrt(3)) = r / 1 = r. tan(pi / 3) = sqrt(3) ==> r = sqrt(3). You original way is over engineered.

@Michael Penn -- more easily to calculate that length=1 line segment at the top: the top tangent segments the hypothenuse into that top small segment, and the longer segment, which you proved it already equals the triangle's base. So, it's hypotenuse MINUS base (4+2sqrt3 - (3+2sqrt3)) = 1

You can get a solution after finding the radius by drawing a horizontal line from the circle center, which ends up drawing a triangle with two of the desired area plus a quarter of the circle

I think every math teacher should watch this.. just so that they can feel how their students feels after explaining something to them.

My man really spent the first 6 minutes to add a one to the bottom leg of that triangle 😂😂😂

I'm glad this was suggested to me, I finished college two years ago but I missed school, especially math class, this was a fun little problem

Very good video, I like your teaching style a lot and I learned a lot from this video. You speak very clearly and explain things well, keep it up.

him: find the shaded area me: *points at the shaded area* FOUND IT!

Went from watching Key and Peele to this, not sure how it happened, but it did.

That was very nice. I really enjoyed it man! I'm almost graduating with a masters in Mech Ing, but I haven't done this kind of calculations in a while, and I really miss it!!

Another amazing video as always; thank you for sharing!

1. From a/b = tan(x) find x = pi/6. Then: c = 2a. 2. Reflect the triangle on its height to get big triangle c-c-2b with circle inscribed. Area of this big triangle: A = a×2b/2 = (c+c+2b)× r/2 and get r. 3. Notice that upper corner has the angle pi/3, thus corresponding sector is pi/6. 4. S = (a - r) × r × sin(pi/6)/2 - pi × r^2/12

HOMEWORK : This is from the Guts Round of the 4th Annual Harvard-MIT November Tournament. Let S be a set of consecutive positive integers such that for any integer n in S, the sum of the digits of n is not a multiple of 11. Determine the largest possible number of elements of S.

That was a beautiful derivation. Thanks!

Didn't get the arithmetic part, but just followed through and accepted it as a given :D Then somewhere half way through the video understood solution, no idea how it wasn't that evident from beginning....Thought I should sub, nice channel...apparently I am subbed already :D Thanks for great, professional and fun video!

This is beautiful

now that im in precalculus, im able to understand this quite well. however, it is still pretty hard and i would hate to have to work this out on my own for an assignment

I saw that 1 from when you got e hypotenuse. The bottom line is 3+2root3, the length of the hypotenuse from the bottom right corner to the tangent point is also 3+2root3, thus making that last little bit 1

That was so sweet, thanks!

Did you know? You are an awesome teacher ❤️

7:33 i saw that peek michael, seems like not even the best of us know the unit circle by heart. don't worry i won't tell anyone ;).

Hi sir. It was an amazing problem. Thank you for the work :)

Nice geometry problem. I enjoyed this. Thank you.

Fun fact: you can do this with the help of co-ordinate geometry .

Love this one!

Nice. Thanks Michael.

13:40 You can also use similar triangles to figure out the angle

You made that much more complicated than necessary. Start by examining theta. The tangent of theta is the ratio of our given sides, which easily simplifies to root(3)/3, (multiply the ratio by [(3 - 2*root(3))/(3 - 2*root(3)] )making our triangle similar to a simple right triangle with sides 1, root(3) and 2, in other words, half of the equilateral triangle. That means the hypotenuse of the original triangle is going to have a length that is twice the length of the smaller side or 2*[2 + root(3)] = 4 + 2*root(3), as you discovered with your complicated technique of “un-nesting” the square root. After drawing the radius perpendicular to our original triangle and determining the two congruent right triangles created by joining the center of the circle with the 30 degree angle, you could see that the hypotenuse of our original triangle is now divided into two segments, and one of the segments is 3 + 2*root(3), so that means the other segment is [4 + 2*root(3)] - [3 + 2*root(3)] = 1. Working with Pi/12 was never necessary. The circular section has an angle of Pi/6 because the small triangle is similar to our original one (they are both right triangles that share the angle at the top). And that quickly reveals the other side and hypotenuse of the small triangle that has side =1 to be root(3) and 2, which shows that the radius of the circle is root(3). We can confirm the hypotenuse of the small triangle is 2 since the original side is 2 + root(3) and the radius is root(3). No complicated trigonometric formula was required. The calculation of the area is then straightforward, just as you describe it.

The right corner point of the triangle, connected to the center of the circle, forms the bisector of the interior angle (the one on the right obviously) of the triangle. I couldn’t ...not say it...sorry. But you could have shown it in a much easier way: considering the fact that radii of the same circle are equal, it follows that the center is equidistant from both sides of the angle, thus the center belongs to the bisector of the angle. Another way is that the two sides of the angle are tangents to the circle. So the bottom side of the triangle and the segment on the right (formed by the right corner point and the point of tangency) have equal lengths.. because ..that’s what tangents do.. lol (Actually, 2 tangents issued from the same exterior point to the same circle are equal). And the line joining this exterior point and the center of the circle is an axis of symmetry of the figure (just the circle and the 2 tangents with no other elements). So by this symmetry, this axis is a bisector of the angle. Also... I prefer to “de-nest” the expressions this way: Let x=radical(28+16radical3) and y=radical(28-16radical3). Then (x+y)^2=x^2+y^2+2xy=64 (just pure calculation) and (x-y)^2=x^2+y^2-2xy=48 which yields that x+y=8 (-8 is rejected since x>y>0) and also x-y=4radical3 (again, -4radical3 is rejected since x>y so x-y>0) adding the above 2 equations gives us 2x=8+4radical3 or x=4+2radical3. I think using the conjugate is better, leaves less room for guess work or trial and error. This gives a really nice idea for a geometry problem for my students. Thanks a lot my friend. Great content.

The hypotenuse can be found much easier by taking 2+sqrt(3) as a and you get the sides to be a and sqrt(3)*a. Hence the third side is 2a. From there on we get, pi/6 directly for theta. After that it should be straightforward to get the area of the triangle-area of sector

I am not good with trig identities and did not feel comfortable working with a pi/12 so my solution diverged around the 10 minute mark. At that point, you can see the top right leg of the little triangle encompassing the area is equal to one and also that the little triangle is similar to the big triangle. That also gives you the radius and therefore the shaded area.

This video should be renamed "How to needlessly overuse trigonometry to solve an elementary geometric problem".

Me as an English major, seeing the first 3 minutes of this… Okay I’ve seen too much KZfaq, need to get back reading my books. 😅

I've always wanted to know the answer to this, but never really tried. Thanks! This math genius reminds me of Pauly Shore.

Thank you very much. Good solution and good presentation.

by the 2nd minute on video i was done applying 30 60 90 triangle and similar triangles he made it far too complicated

That´s the kind of problem you solve in a test only if you have memorized the solution the day before.

Thanks, I appreciate and enjoy.

youtube recomend me this video 6 years late but i finally know how was the way of solving this problems

Michael, big fan here. But this was needlessly complex. It is easily seen that both tangents on the circle of the yellow triangle are the same length. You even said so in the beginning. Thus the length of that side of the small upper triangle is one. (4+sqrt3)-(3+sqrt3). The vertical side of the small triangle is (2+sqrt3-r). Some simple math shows that r=sqrt3. And then the rest is as you explained.

Ignore the nit-pickers. It is interesting to see the many ways we could potentially navigate such a problem. Often for mathematicians, knowing the vast array of tools you have at your disposal is the most important thing. Knowing that one thing can be expressed another way, that you can reveal information about A by doing B first, that the more you dissect the given information you have, the more you have to work with, which makes it easier to solve the problem, is ultimately what this channel is about.

This is quite impressive. It's quite obvious what you did in hindsight, but even though I know the math behind this puzzle, I wouldn't have been able to put this together start to end.

You choosing not to just solve the simultaneous equation by substitution was killing me lmao. So much more reliable