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See, that’s what the app is perfect for.

Sounds perfect Wahhhh, I don’t wanna

Take a triangle whose sides have lengths that are integers. Then compute the sine and cosine values of the angles and compute the area. Something surprising happens! The area and the sine values have a similar form, they are a rational number times the square root of an integer, and the cosine value is always rational! 

Try to prove that this always works and that you get the same number bellow the square root sign if you take every thing out that you can. For example getting sqrt(12/49)  and sqrt(3) is fine since  sqrt(12/49)=sqrt(2*2*3/7*7)=2sqrt(3)/7, so booth values are rational multiples of sqrt(3). Also getting sqrt(15) and is sqrt(5/3) is fine since sqrt(5/3)= sqrt(5*3/9)= sqrt(5*3)/3= sqrt(15)/3.       

math theorem residual of a triangle triangle
I was quite surprised the other day. I was doing some research on odd-distance graphs, and suddenly this drawing popped up. I knew that the points will lie on a circle but I didn’t expect them to be so regularly spaced. This is almost a regular...

I was quite surprised the other day. I was doing some research on odd-distance graphs, and suddenly this drawing popped up. I knew that the points will lie on a circle but I didn’t expect them to be so regularly spaced. This is almost a regular 33-gon! Which is interesting since regular 33-gons cannot be constructed with ruler and compass, but this drawing can be constructed.

 This how it was created: I took a circle of radius 7/sqrt(3). Then I picked a point, it is denoted by (0,0,0) on the image. Then I started to add edges from that point of length 3, 5 or 7. So each segment that you can see have length 3, 5 or 7. And I did this recursively, from the new points I added more segments of length 3, 5, or 7. The triples on the points denote how it can be reached from (0,0,0). If (i,j,k) is written on a point then you can reach it by drawing i segments of length 7 and then j segments of length 5 and then k segments of length 3. You can see that some triples are missing, for example there is no (1,0,0) on the image. The reason for this that some points can be reach in multiple ways. For example (1,0,0) is the same as (0,1,1). (To see this just calculate the radius of the circumcircle of a triangle with side lengths 3, 5, 7. You will get 7/sqrt(3) ) Then I generated all (i,j,k) points where i,j,k<6 and this was the result.          

(If you are interested in these kind of odd-distance drawings you can read my paper for a nice theorem: https://arxiv.org/pdf/2008.10305.pdf)

math theorem art odd-distances