Hi, bprp. It's Cameron with another challenge for you. Is there a function f(x) such that the integral from a to b of f(x) is b-a for any positive integers a and b (assuming a

I'd like to see the same problem but with a circle. of course not dividing a circle into 2 pieces, that's too easy, so how about dividing a circle into 3 equal pieces using vertical cuts

Woo! When I was young, I used to play around questions like this. Shape and ratio stuffs. Time flies. Bring me nostalgia. Thank you BPRP.

I forget how much I enjoyed algebra until I bump into something like this.

Once you realise that the area is halved, you can scale the triangle down by a factor of 1/√2, so the height will be a/√2 It's quicker and easier

I love that I would have approached this completely differently, to the same result (A1 is a rectangle b1*h, plus a similar triangle). This is simpler though

I’ve never seen someone so passionate about math. Keep it up, I like the videos.

A nice and easy problem and a cool teacher always make a great video ! Keep going

What a delightful problem! Very satisfying to see that solution come out so clean and simple.

I used the similarity of the two right triangles. When two triangles are similar then the ratio of their areas is equal to the square of the ratio of the corresponding sides. And we have the ratio of the area of the areas of the two right triangles (1/2).

i love this channel so much

Very elegant solution. Thanks for posting!

Very nicely done. Thankyou

Great video!

I love solving for surface area of 3D shapes it’s so satisfying to do and I liked the Pythagorean therem for right triangles so satisfying.

The area of any irregular shape scales in proportion to the square of its length. So for the triangle we are looking at: b² = 2b₂² b₂ = b/√2 This does not just apply to triangles, but any squiggly shape that you draw. Any irregular volume scales in proportion to the cube of its length. This makes perfect sense if you think about it. Remember this - it will help you to easily solve many problem that you come across.

Lovely. Thank you

Just found this video and found it kind of interesting. A variation on this I was working on was to find a line, coming out of the right angle up and to the right with angle theta, would also divide the triangle in two equal halves. Seems like a daunting problem, since I had 7 equations with 7 unknowns, but I think the answer ends up being tan (theta) = a/b.

Using scales: length scale k=b2/b -> area scale k^2 = (b2/b)^2 = A1/(A1+A2) = ...= 1/2 -> b2 = b*sqrt(2)/2. Thank u!

Cool question. Nice answer.

Small triangle similar to big triangle and ratio of areas is 1:2 So ratio of sides is 1: rt(2) So ratio of sides of trapezium and big triangle is (rt(2) - 1)/rt(2) So ratio of sides of sm triangle and trap = 1:(rt2 - 1)

you can use triangle similarity on the two triangles to get immediately that (b1/b)^2=1/2 => b1 = b/sqrt(2)

just for fun I did some algebra and came to the conclusing that b2^2 = b^2/2 and for fun wanted to construct the length on b (which is easy but why not), conscruct a square, divide the 4 sides in 2, connect the midpoints and now you have a new square, put this square's length on the vertex and conscruct a line paralel to the oposite side, now you have divided the triangle in two equal parts

What a satisfying solution

Amazing video (:

Amazing content. I would love to see a video for all triangles not just 90 degree ones. Also how could you split a convex shape in a half. given by any amount of points in cartesian coordinates, i mean get formula of line splitting it or just the points in which it intersects with shape.

Thank you bprp for geometry challenge! Bring back so much memory 😁 How to divide a right-angled triangle into two equal parts in one second? Join the vertex of the right angle to the mid-point of the hypotenuse 😁✌

I started off like "That's suspiciously root 2-ish"

I solved by vertically stretching the triangle so that it’s hypotenuse draws y=2x and then integrated for half the area using x^2

Hang the triangle by a corner, and draw a vertical line. Repeat with another corner. Draw a perpendicular at the intersection to any edge. Voilà!

You can also do this with Calculus! Consider the line y=ax/b and take two integrals, integrating the first from 0 to t, and the second from t to b. These two integrals must be equal since they give the area under the curve and we want the areas to be equal. After integrating and solving for t, you get the same answer.

I hope I wasn't the only one who solved this using the trapezoids and the smaller triangles areas 😞

Awesome. I dont know if i will ever use this, but i hope to one day

does the midpoint of the hypo connected to the right angle works?

think of the line labeled 'a' as being on the Y axis and the line labeled 'b' as being on the X axis of a graph. You then have a simple linear equation for the diagonal. if you work it out you get: f(x) = -(a/b)x + a Then you want to find the variable x_1 so that the integral from 0 to x_1 of f(x) dx = 1/2 * integral from 0 to b of f(x) dx. That way A1 is equal to half of A1 + A2, i.e. A1 = A2 we know that: F(x) = -(a/2b)x² + ax + C So you get the equation for x_1 which is -(a/2b)(x_1)² + a(x_1) = 1/2 (-(a/2b)*b² + ab This forms a quadratic equation that you can (simplify and) solve for x_1 in terms of a and b

Hey dude, I'm 8th grade taking Math 1, I found this super interesting! I think you should try a hendecagon into 3 equal pieces next!

I love math very much Sir literally I enjoy solving maths it's very fun to solve maths

Please explain method of difference in AP series

Nice! I saw a similar solution given to similar question using a cone. Pointed side down, at what height do you cut so that the volume is half of the original? You multiply the height by the cube root of 1/2 to get the answer. Saw it in a ted talk.

A line passing through two line segments of a triangle, parallel to the third side divides them proprtionally (in the same *ratio* )

What about for an arbitrary triangle, as opposed to a right triangle? Where to make a cut on any side of any triangle perpendicular to the side (although not necessarily through the side) such that the resulting shapes have the same area?

Nice Video.

Anyone else use compass and ruler? Edit: I realised one would use them to draw the line, but drawing it doesn't justify the sq root 2. The parallel line makes a smaller, similar triangle, and a side-length ratio of sq root 2 makes the area of the smaller triangle half the area of the original triangle, and thus the remaining area is also half.

If you bisect the hypotenuse of a right triangle and go out either direction from that point. The area subtended by the (x,y) coordinates of either location (equidistant from the bisection) will have equal areas. That is x1 × y1 = x2 × y2.

To get the answer I'd ask the magic 8 ball you're holding in your hands

I set the area of a trapezoid equal to the area of a triangle, then just solved for the base of the trapezoid. It took a lot more math than used in the video, but it was definitely really fun. I had to complete a square but with negative rectangles getting added on. It was kinda wack.

Similar triangles, small = 1/2 big, so 2cd = ab and a/c = b/d. In other words d/b = a/c = 1/√2

For any triangle: draw a line that pases through the centroid

Write b_2 as a fraction of b, i.e. b_2 = b/r (and so b_1 = b(1-1/r)). Then h = a/r, so A = ab/2r^2 = A/r^2. To get A/2, we need r = sqrt(2).

Best punchline ever '' just for you guys ''

Id like to see it but with the shortest line possible

I used integration, and treated the hypotenuse as the reflected function y(x) = (a/b)*x.

How can I measure the radius of a circle inscribed in an isosceles triangle as a function of its smallest side?

a*b = 2h*b2 Since h/b2 proportional to a/b for some k, a*b = k*h*k*b2 But k*h*k*b2 = 2h*b2 implies k^2 = 2 so k = root 2

想看你解一下這一題，過直角三角形斜邊一點D作一條垂線，使該垂線平分三角形面積，試以兩股長表示D點到直角的距離。

I like this solution. It's always beautiful whenever a geometry problem like this can be solved using the principles of geometry alone. I solved this in a different way, using calculus. Let y=-a/b*x+a, then let A=S_0^b y dx, and find b_1 such that S_{b_1}^b y dx=1/2*A. We have A=S_0^b -a/b*x+a dx=-a/2b*x^2+ax|_0^b=-ab/2+ab=ab/2 and S_{b_1}^b -a/b*x+a dx=-a/2b*x^2+ax|_{b_1}^b=-ab/2+ab-(-a/2b*b_1^2+ab_1)=ab/2+a/2b*b_1^2-ab_1. Setting S_{b_1}^b y dx=1/2*A gives: ab/2+a/2b*b_1^2-ab_1=ab/4 ab/4+a/2b*b_1^2-ab_1=0 b/4+b_1^2/2b-b_1=0 b_1^2-2b*b_1+b^2/2=0 b_1=(2b+sqrt(4b^2-2b^2))/2=b+1/2*sqrt(2b^2)=b+b/2*sqrt(2)=(1+sqrt(2)/2)*b=(1-sqrt(2)/2)*b since b_1

It´s just beautiful.

this is really cool.

I solved it through integration and once the answer was in front of me I realised immediately I had taken a very long way around. A slegdehammer to crack a nut!

A quicker solution could be that, since the 2 triangles are similar (because of Talete's theorem), all sides are increased by the same ratio. This means that we want to solve for the ratio x: (ab/2)/2=axbx/2 where ab/2 is the area of the large triangle and axbx/2 the area of the small one. ab/4=abx²/2 x²/2=1/4 x²=1/2 x=1/sqrt(2) Therefore, since the sides of the big triangle are a, b, c=sqrt(a²+b²), the sides of the small one will be ax=a/sqrt(2), bx=b/sqrt(2), cx=sqrt(a²+b²)/sqrt(2)

I tried to solve before looking at the whole video and I have got the answer z = ((sqrt(2)-1)/sqrt(2))b, this comes up to around 0.2928b. Where x=z is the line that cuts the triangles! Used Integration to do it! Rather easy to do. Verified on desmos!

I don't know why we all complicated it so much (I tried different methods and even got lost with several equations); now that I see the solution, it's rather obvious: if you need half the area with a proportional shape, the sides must be in proportion 1:√2.

@blackpenredpen For any triangle if we a cut parall to 1 side at a distance of 1/sqrt(2) from opposite vertex would give the half area portions

Hey BPRP ! You could call that the magic cut.

Hey BPRP! This is very interesting to me. It's interesting that the length of a doesn't matter.

let's see: the divide is x units to the left of the point. The resulting area is (x/b)^2 times the original area (by proportionality). we want (x/b)^2 = 1/2, easy. x = b/sqrt(2)

Divide an ellipse into equal areas?

Perfect for you ❤️

Hi, Here is a challenge: sqrt(x)=-1 the sol is not to square both side, because if it was the sol 1=-1 hint: log_(sqrt(x)) both side

Can you explain all the ideas using more writing because I do not fully understand English

in a square or rectangle, b1=b2 = b/2 and u get the same are In triangle is b/sqrt(2)

Obviously, we want ( b2/b)**2 = 1/2 (similarity of triangles)

blackpenredpen what is your real name??? I like your content on calculus...your style of teaching is awesome.

Or more simply: If A2 is half of A, and it's a similar triangle scaled in 2d, then if the area is to be halved, the lengths have to be divided by root2

I love that you are so responsive and connected to your audience

Can someone clarify why my thinking is wrong. The centroid (center of mass/geometric center) of a right angled triangle is 1/3*b, so why isnt it the case with this problem?

Hey, bprp. It's Cameron again. The challenge is this. If you get out your calculator and enter cos(0), of course you get 1. But if you then enter cos(ANS), and keep pressing the equals button, then it will converge around a constant, around 0.798. Can you give this constant in exact form? (Note: of course, the decimal places go on forever, so you can't just list it. So, give it in some simple form eg. the golden ratio is (1+sqrt(5))/2.)

曹老師 In the intro and 2:23, what happened? Was a student going to talk to you?

Plus court et avec tes notations. Avec b2 qui se lit "b indice 2" et rac(2) qui se lit "racine carrée de 2" A = 2.A2 donc 1/2(ab)=1/2(2.hb2), donc ab = 2.hb2. De plus A et A2 sont des triangles SEMBLABLES, donc il existe k tel que a=kh et b=kb2. En remplaçant: k^2.hb2 = 2.hb2, en simplifiant: k^2 = 2 donc k=rac(2). Pour finir h = a/rac(2) et b2 = b/rac(2). Amitiés

For this one it's easier to work explicitly woth the scaling factor k. Replace a2b2 with kakb and get kk=1/2 after cancelling. Juggling cross ratios is kind of distracting here.

Anyone realised that he switched the markers impressively?

I find it interesting that the length of b1 or b2 is not dependent on a, which means it is not dependent on the angle. It seems bizzarre(in other words counter intuitive)

Somebody's trying to mess w/ you? Tell them off professor!

Hi i have a series problem that i have been solving but the answer is awalys 1 that i am getting... summation from -inf to +inf of sinc^2(k/a)...the question is asking to prove this to be a and i am getting 1 always

I would rather use similar figure concept -> b:b2= sqrt(2):1, problem solved!

Geometry! Yay!!!😃🤓😃

h = a / sqrt(2) Because of the similar triangles.

There is another way to do this. There is a theorem that says if too triangles are similar then the ratio of their areas is equal to the ratio of the similarity squared I learned this at 11th grade.(which i am still in). A=2A2 so A/A2=2 so (let's call the similarity ratio r) r²=2 so r=✓2 which means that a/h=✓2 so a=✓2h and h= a/✓2. Then we know that A=1/2 ab and that A2= 1/2 h*b2 and also that A=2A2 so we find that b2= ✓2b/2 and then if we subtract b2 from b we can find b1 as well!

What about a formula that can work for a cut in any direction?

its werry funyy

Chose two random pair of points in a square. Draw a straight line between each pair. How high is probability of the two lines intersecting?

a quicky : how many equilateral triangles can be inserted in a circle at max? assuming 1 side of an equilateral triangle is equal to 1/4 of the radius

To me, sum of all natural numbers = -1/12 doesn't make sense can you help me "VISUALIZE" It???

Everyone: geometry Me, an intellectual: the triangle can be defined with a slope of a/b, and treating that as a line let's us integrate it to find the area of the triangle as a function of x(or b), Obviously from that point onwards we're just looking for what value of x makes the area half of the max(at x=b where b represents the total base of the triangle)

How to give you challenge question??

Yo this is like how A4 length is twice of A5 breadth, and so on for A5 to A0, because the ratio of length to breadth is sqrt(2)

Here's a much simpler way: Assume the two acute angles of the triangle are 45 degrees. This will not affect the location of the cut, as we can scale the triangle vertically up and down to get any shape right triangle we like, and the two areas will keep the same proportion. This means we're working with an isosceles right triangle. Let's further assume each leg measures 1 unit (we can scale this to any size we like at the end). After the cut is made, call the height of the smaller triangle h (which is also the other leg, since the small triangle is also isosceles). The large triangle has area 1/2. So the small triangle must have area 1/4. Thus h²/2 = 1/4; h² = 1/2, and so h = 1/√2 (also called b2 in the video). Since we assumed the base to be 1, multiply this result by the actual base length (called b in the video) to get b/√2 (b2 in the video). Of course we can calculate b1 the same way you did it towards the end of the video; that is, b - b/√2.

what about a cut thai isn't parallel at a?

Doesn't the similarity imply more than this? Namely, not only that b_2 equals b times a scaling factor, but also that h equals a times that same factor (let's call it r)? I. e. (1/2)h(b_2) = (1/2)abr^2 = (1/4)ab , so r^2 = 1/2 .

I immediately went root 2, because it's the integral of a slope so it's quadratic.