Question number 23, the diagram shows several non-overlapping isosceles triangles with a common vertex at O. Every triangle shares an edge with each immediate neighbor. The smallest angle of a triangle with vertex at O has angle measure M, where M is a positive integer. The other triangles have angles at O of measure 2M, 3M, 4M, 5M, and so forth. Here, the diagram shows an arrangement of five such triangles. The question is, what is the smallest value of M for which such a set of triangles exists? So let's fill in the information we are given. The smallest angle measure we have measures M degrees, then the next largest angle with vertex at O is 2M, 3M is next, 4M, and with five triangles, 5M is the largest angle measure of an angle with vertex at O. And we note that the sum of those angles have to add up to 360 degrees. So in general, with K such triangles, 360 degrees will have to be the sum of M plus 2M plus 3M all the way up to K times M. And we can factor out the common factor M. And what we have left over is the sum of the first K positive integers, which simplifies as the product K times K plus 1 divided by 2. So now we have 720 is equal to M times K times K plus 1. And our job is to minimize the value of M. And that is the same thing as to find the largest possible value of K and K plus 1, positive consecutive integers in the factorization of 720. So let's factor 720. And what we see is that 720 is equal to 8 times 9 times 10. That comes out to 2 to the power of 3, 3 squared times 2 times 5. And the largest consecutive factors we can write down are 15 and 16. So those must be the values of K and K plus 1. And with that, we are left with 3 as the value of M that is the least possible. And that is the answer to number 23.

isosceles triangles

smallest integer M

vertex angles

angle summation

minimum angle