Mate, you rock. There are some channels that would have taken 15 minutes to explain that, and it still wouldn't have been as easy to understand as your explanation.
The area of a flat figure circumscribed around a circle is equal to half of its perimeter multiplied by the radius of the circle. For the right-angled triangle with sides 3, 4, and 5, the area is calculated as 3*4/2=6, and the perimeter is 3+4+5=12. Half of the perimeter is 6, so the radius of the circle is 1.
I am 61. The fact that you shared is absolutely remarkable to me, but even more remarkable that I have somehow never run into this before. I work in the math department of a local high school, and I will be sharing this with my coworkers next week. And I will be subscribing to your channel.
When they raided Einstien's office after he died, they did not find what they were looking for. Stven Hawkins did not work it out either, but it did concern Pi.
I'm 50 How is that possible that you've never come across it. i learned this back in high school with 1 difference, we didn't call it a 3 4 5 triangle but other than that the concept is the same but did you , a math teacher , not know this since you are older than me
What a great explanation. Simply done in such a short time. I'm 73, a retired engineer. It's so good to see a young person with such a special explanatory skill. I've subscribed!
For any rationally scaled version of a 3-4-5 triangle, the area divided by pi is then rational and a perfect square. Clarification: The area of the inscribed circle is what I mean.
That was so much fun I ended up with a big grin on my face! I love the 3-4-5 triangle, and never knew this added feature of the a circle inside, and its area. Well done!
On the ranch,back in the 70's a brilliant mathematician showed me this triangle when I was building fences. thank you, I have now learned the mathematics of it. I always made straight fences, rooms, etc. I made a wooden tool of this dimension 50 y ago and passed it on to my grandson.
If you know the generators of the triangle, you can find the radius of the incircle. it is simply (a+b-c)/2, so (3+4-5) / 2 =1 So the 5, 12, 13, which gets you 2 as a radius And finally the last of the regular 3 you see on tests; 8, 15, 17: has a 3 radius And yes you can prove all the radius of right triangles with integer values will always be a integer radius.
@@clickrick This only works for right triangles. If you take a triangle such as a 4,4,4 which is an equilateral triangle with integer coefficients you will not get the same results.
@@Lord_Volkner Given your other post about squaring the circle I very much doubt that you would understand the proof. Anyway, in the video above replace 3,4 and 5 by x,y and z and out will pop your proof.
The general form of a Pythagorean triple has sides of length k(m^2 - n^2), k(2mn), k(m^2 + n^2), for m > n > 0 and any k > 0. The radius of the circle inscribed in such a triangle is r = (m-n)nk. Which is clearly a whole number for any Pythagorean triple.
@@any2xml set k = 1, m = 2, n = 1. Then the sides of the triangle are 1(2^2 - 1^1) = 3, 1(2*2*1) = 4, 1(2^2 + 1^2) = 5 and the radius is r = (2 - 1)*1*1 = 1
I’m 56, a bricklayer by trade. Yep, I grew up in an olden days. We walked onsite with a basic plan and very often set out a house foundation and the old 3-4-5 was never left out. We knew no other way, certainly had no other tools at our disposal to do it any other way, and why would we?🤷♂️. It was always satisfying to be the one(s) to set in concrete, brick, 🧱, the building to be.
Nice example! I want to point out that the incircle radius of a triangle is always r = A/s, where s is the half perimeter of the triangle and A is the area. This relationship is not so often taught in school. And it don't even need to be a right angled triangle! In this example, we have s = (a+b+c)/2 = (3+4+5)/2 = 12/2 = 6. Area is 3·4/2 = 6. Therefore we get r = A/s = 6/6 = 1, and πr² = π.
Interesting, but calling the perimeter of a triangle (or any not circular shape) it's "circumference" is not standard terminology at least for myself so that was confusing.
Yes that's exactly how I learned it at my school. We have a lot of exercises using this formula (although we write it as S = pr, where S is the surface area, lowercase p is the semiperimeter, and lowercase r is the inradius)
In fact, any Pythagorean triple (or multiples thereof) will have an inscribed circle with an integer radius. Cf. 5-12-13 or 7-24-25, e.g. More generally, if sides a, b and c are rational, the inscribed circle's radius will also be rational.
In farming, for palning new acres (or in my case new wineyards) we used GPS (of course) but we also learned the oldscholl way, using the "12-knot-string" (3+4+5 is 12). Very handy tool! As long as the distances between all knots is equal, it doesn't matter how long the rope or the distance between the knots really is. Using an about 20m long rope gives about 1,5m distance between the knots, its accuracy was astonnishing!
Okay, Andy, this is good actionable information. Just what I needed. Thank you! I'm a math enthusiast myself. And what I would like to know is what is the area of the other three sections within the triangle. Are all three of the sections irrational numbers? Do they have any magical properties when compared to one another? I am dying to know.
Very neat! I have a doctorate in maths and have never seen this (that I recall). It gets better. The other two most common small triangles with Pythagorean triples are 5, 12, 13 and 8, 15, 17. As the "r" in those cases is 2 and 3, the inscribed circle areas are, respectively, 4π and 9π. A general equation for a triple (a,b,c) is easy enough to find (exercise for the reader!). The circumscribed circle is sadly rather dull.
Hi Andy, you have presented a well-illustrated article that displays a result not well known (judging from the comments). It is worth noting, though, that for any integer r, a Pythagorean triangle can be constructed with sides of integral length and having an inscribed circle of radius r with area r2. Obviously, this can be done, trivially, by taking all multiples of 3,4,5. However it can also be achieved with combinations of sides, many of which are relatively prime, as follows. For odd r: side lengths of r2 + 2r, 2r + 2, r2 + 2r + 2. The first three in this sequence are: 3, 4, 5; 15, 8, 17 and 35, 12, 37. For even r: r2/4 + 2r, 2r + 8, r2/4 + 2r + 8. The first three in this sequence are: 5,12,13; 12,16,20 (a multiple of 3,4,5) and 21, 20, 29. The first formula can also produce side combinations for even values of r, but many more of these are multiples of previous results.
That was a brilliant observation. I've never noticed such a beauty in the 3-4-5 triangle. What's even more fascinating is that the area of the incircle is π = 3.14159... *Notice how the numbers* _3, 4, 5_ *pop up in π, one after another.* And what are the dimensions of the triangle? _3, 4, 5_ . Majestic! *_• Constructive Criticism_* This method was a unique one, using properties of tangents. Another method will be to join the incentre with the 3 vertices. This will divide the triangle in 3 smaller triangles. Now, the 3 sides of the triangle are tangents to the incircle. Hence, the 3 inradii at the 3 points of contacts will be perpendicular to the 3 sides. You can then use the formula of area of a triangle on the 3 smaller triangles and add them up and equate the sum to the area of the larger triangle. This will actually derive you a shortcut: Δ = r · s (where Δ = area of triangle, r = inradius, s = semi-perimeter of triangle) _# Great approach and observation nonetheless._
The area of a flat figure circumscribed around a circle is equal to half of its perimeter multiplied by the radius of the circle. For the right-angled triangle with sides 3, 4, and 5, the area is calculated as 3*4/2=6, and the perimeter is 3+4+5=12. Half of the perimeter is 6, so the radius of the circle is 1.
ALL Pythagorean triples (right-angled triangles whose sides have whole number lengths) have inscribed circles whose radii have whole number lengths. This means that the area of such circles will always be a whole multiple of pi. Further, if the triple is (a, b, c), then r = (a + b - c)/2 and r = (ab)/(a + b + c). Equating these expressions for r leads to a^2 + b^2 = c^2 (Pythagoras' Theorem). Of further (possible) interest is that the two expressions for r can be stated as: half the difference between the hypotenuse and the sum of the legs, or the product of the legs divided by the perimeter.
Pretty cool. But by saying, "This is true for every 3-4-5 triangle," tou seem to be either stating the obvious--that all 3-4-5 triangles are the same, or that there exist multiple unique 3-4-5 triangles.
Totally agree that this is worthy of note! I had no idea. More evidence that the physics and geometry of our universe is often amazingly simple at its core.
Well, the video only shows it to be true for Euclidean geometry. But in the physical world, Euclidian geometry only happens to be approximately correct in areas were spacetime curvature is negligible. So no, this does not provide any evidence that the physics and geometry of our universe is amazingly simple. Physics require empirical observation, which there are no examples of in this video. Such observations reveal that the geometry of our universe is more generally Lorentzian, not Euclidean.
Awesome. How you solved for the radius, was easy to understand, even if you had basic algebra. I often use the 3-4-5 rule while squaring up things I'm building. That this rule works, is an understatement. That said, learning that an inscribed circle, tangent to the sides, within this triangle, will equal pi, is a wild thing to learn. The same for the lengths of the tangent lines. Too bad they don't (didn't) teach this in any geometry or trig class I took.
YES! Despite my frustrations where math is involved (just me being bad at it while I want to be good at it) I only have admiration for something like this. It has natural beauty. This also feels like it has deeper meaning.
It's the Pi because the radius is 1. But the "Circle tangent to the three sides" problem also has the 2nd solution: R=6 and area=36*Pi. The circle will be tangent to the "5" side from the outside and to the "3" and "4" sides on their extensions beyond the triangle
It's interesting that the area is pi, but that just comes from the somewhat less exciting fact that the radius is 1. As other commenters have alluded to, there's an entire class of triangles for which this is true. One way of generalizing them is that this will be true for every triangle (right or otherwise) whose area is half its perimeter. (The 3-4-5 triangle has area 6 and perimeter 12.)
Not sure if this was already stated in the comments below, but this same proof can be used to show that the area of the inscribed circle is an integer multiple of Pi for all Pythagorean triple triangles. Pretty cool.
That's really awesome! Thanks for showing us that geometric phenomenon. When you say that this works for "all types of 3-4-5 triangle", aside from the one you've shown, what other types are there?
There aren't any others aside from drawing the triangle sideways or backwards or rotated. To get r = 1 from integer sides you need a = 3, b = 4, c = 5. But actually something extremely similar happens for _any_ right triangle. The radius of the inscribed circle is always r = (a+b-c)/2.
Simply ASTOUNDING and right under everyone's nose, as in FOREVER. I can't help but think of the countless teachers and scholars worldwide, including myself, unaware of this beautiful little proof. Can you cite a source for this proof or share a little about how you learned it?
Very nice, like all the videos in this channel! But I see this more as a riddle than a math question. This is based on the property of the inscribed circle's radius to be equal to 1/3, 1/4 and 1/5 of 3-4-5 triangle sides. Whenever you consider a circle to has a one-unit radius, you will always have a circle with an area equal to pi. In such case, the "trick" is naming such triangle as 3-4-5. On the other hand, if you name it a 6-8-10 triangle (technically also a 3-4-5 one), the resolution won't fit to pi. It's a good one!
It would be double pi since the radius of the circle would be 2 instead of 1 when we double the lengths by making them 6 8 10 instead of 3 4 5@@letlhogonolokebasitile6924 Edit: I meant square of pi instead of double
I can’t really see what point you’re making. It doesn’t matter what length you call the sides. It’s only the ratios that count. For any 3-4-5 triangle the ratio of the area of the circle to the area of the triangle will be pi/6, and that’s what counts.
Could it be that this works for all pythagorean triples? 🤔 EDIT: Wait, that doesnt make any kind of sense, after I started drawing it out 😂 But maybe the ratio between areas at least are the same? EDIT 2: OK, so far I have worked out that the radius seems to always be a whole number. First I thought that the radius was 1 unit longer for every pythagorean triple, but both the 7,24,25 and the 8,15,17 triangles has an inscribed circle with a radius of 3. The 5,12,13 triangle has an inscribed circle radius 2. Next up is 9,40,41 and sure enough the radius here is 4.
Let (a,b,c) be a Pythagorean triple. By the method set out in the video, we get: c = a - r + b - r c = a + b - 2r 2r = a + b - c r = (a + b - c)/2 So the area of the inscribed circle is 𝝿 * ((a + b - c)/2)^2. Given there are infinitely many Pythagorean triples, can we prove the area will always be an integer multiple of 𝝿? Will r always be an integer?
@@colinslant Yup, so the last step is to see if (a+b-c)/2 is going to be an integer value, or to prove that (a+b-c) is an even number. Pythagorean triples are always three even numbers or two odds numbers and one even number. Both of these cases will result in an even number, hence we should expect an integer multiple of pi
@@curtishuang5534 But we need a proof of that last statement! A Pythagorean triple can obviously consist of three even numbers, since if you multiply the members of any triple by a power of two the result will be a Pythagorean triple containing three even numbers. Is there a proof the only other possibility is two odds and one even? It's probably in Euclid...
Like everyone else, I was very surprised by Andy's discovery and his beautiful demonstration. I now see that the same conclusion could have been reached in another way. Around a circle of area π we can construct an infinite number of right triangles with legs A, B not necessarily integers. We could call them π-triangles. Applying to these triangles the same considerations used here by Andy, it is easy to show that their hypotenuse is A+B-2 (so the 3-4-5 triangle is a π-triangle, since 5 = 3+4-2) and that their area is A+B-1 (so the triangle 3-4-5 is a π-triangle, because its area is 6 = 3+4-1). Those two general properties lead to a curious corollary: in π-triangles, the perimeter is twice the area. Again, the triangle 3-4-5 is then a π-triangle, since 3+4+5 = 2x6.
I liked math since before I started school (I am 71 now). Worked a lot with math - never saw this before. Elegant and easy to understand. The only thing that disturbs me is my perception - to me the area of the circle "looks like" it smaller than the rest of the triangle (the area of the triangle is obviously 6 so the area of the circle is slightly more than half of the rest). This must be due to a delusion or misinterpretation that I suffer from. Wonder if anyone else got this feeling. I trust math more than my perception which in this case seems to fool me somewhat.
There is a nice geometric way to solve. Make 4 copies of the triangle, each one with a different 90 degree rotation, which would put the hypotenuses as such: upper right, upper left, lower right, lower left. Place them together to form a 5x5 square. That would make each side length of 3 share a side with a side length of 4, and arranging this way puts a small square right in the center of the 5x5. Now the area of the 5x5 is 25 sq. units. The total area of the 4 triangles is 24 sq units. So the area of the central square is (25-24) sq. units, or 1 square unit, and being a perfect square, it has a side length of 1, which is the same as the radius of the incircle of the 3,4,5.
Amazing symmetry. Geometry gets used more than some people think. In my machine shop classes we would have to try and calculate the radius tangent points of a cutting tool to determine perfect tool path and direction. In order to get the desired radius or angle in the metal parts. It got really complex. Fortunately with computer controls, the machines can actually do the micro trigonometry for each individual tool, and we only have to input required dimensions. But understanding the principles helped if something wasn’t running through the metal as needed.
3-4-5 triangles were a bit of a dead end in trigonometry. So for any triangle, there will always be a scalar modification that has the unit circle tgat fits inside of it. It simply means you can scale triangles. 3-4-5 traingles were developed by the babylonians, see plimpton 322 tablet, in order for the construction of rectlinear structures with few nonirthogonal angles. The gate of Ishtar and the Ramp of the Ziggurat of Ur are examples. The problem with the these triangles is that you cannot build circles with them. The only types of triangles that can be used directly to construct circles are the the isosceles triangles, for the unit circle the angle needs to be an integer when dividing 2pi (or 360). These triangles of the unit have a single unique quality, the chord. 120 - SQRT(3) - Triangle 90 - SQRT(2) - Square 72 - SQRT(1-(golden ratio/2)^2) - Pentagon 60 - 1 - Hexagon 51.4 - 0.87 - Heotagon. The only isoceles right triangle is the 90 -45-45. Which the babylonians solved the chord to 6 digit places and pythagoras cult discovered it was a radical. This is possibly the first chord solves. Any unit circle can fit roughly into any polygon but scale of the triangle must be a function of 1/(bisector of the chord). If you take a equilateral triangle, for instance the and you want to inscribe a circle into the triangle. The inscribed triangle has a chord of SQRT(3). The half chord is SQRT(3/4) and since its a unit circle the SQRT(1/4) = 1/2. This means a unit circle will fit into a triangle of scale factor 2, or sides of 2SQRT(3). This means each side is ~3.5. This means the circle fits much more effectively into an equilateral triangle than a 3-4-5 triangle. More over, what if I wanted a structure with integer sides. The inscribed square has sides chord of SQRT(2). If we wanted to inscribe the unit circle into a square, we need to divide it by the 90° bisector which is SQRT(2)/2. Thus give side lengths of 2. Square has combined side lengths of 8, the 3-4-5 triangle has side lengths of 12. The hexagon of sides 2/SQRT(3) encloses a unit circle. At six sides 2*SQRT(3)^2*2/SQRT(3) = 4 * SQRT(3) ~ 7 units. If we continue this process of fitting as the we increase the sides of a equilateral polygon eventually the length of thr sum of the chords is mathematically indistinguishable from the sum length of chords of the inscribed scalar polygon which is indistinguishable from pi. The real break through in trigonometry is the identification of nearly isoceles "perfect" right triangles. This allowed the determination of SQRT(2) to 5th decimal place (base 60 equivilent thereof). It is from these triangles that pythagoras used to formulate his theorem which basically allowed the creation of chord tables, the predecessors to the sins and cosines. Sin x = 1/2 chord 2x Cos x = bisector of chord of 2x So once you have all chords from 0 to 180, you have sin and cosine for 0 to 90. Thank the greeks for wanting to build circular temples instead of square ones.
Excellent explanation. I started drawing radii of the inscribed circle but took a while to realize the tangent segments theorem. Really cool stuff man.
Taking into account its inscribed circle, I always like to point out that the right triangle 3, 4, 5 is in fact a 1 (radius of the inscribed circle), 2 (its diameter), 3, 4, 5. Note that if we extend this triangle by forming the rectangle of length 4 and width 3, the length of the segment between the 2 centers of the inscribed circles is root of 5.
This is the backwards in Time triangle that Dr. Sabine (and Dr. Von Hauser who was actually real and you can't get his work in normal channels) was talking about. Observer position and dealing with getting somewhere before you left due to polar elements. Thanks.
It's not that hard to derive the equality r² = (c-a)(c-b)/2, which holds for any right triangle. (The general case is a bit more complicated but still manageable.) Now a = 3, b = 4 and c = 5, so r² = 2*1/2 = 1, that is r = 1.
Fun fact: there are 16 other primitive Pythagorean triples (aside from obvious multiples), with integers under 100... ( 3, 4, 5) is just the first one. My fav is 11,60,61
