Hey guys so i have an impossible problem that needs to be solved. So this equilateral triangle's sides are divided into three equal parts. There are lines drawn from the corners of the triangle to the one of the parts divided. Those lines then create a smaller equilateral triangle (shaded area). What is the ratio of the big equilateral triangle to the small equilateral triangle?
An equilateral triangle's sides are divided into three equal parts.
There are lines drawn from the vertices to one of the divisions.
Those lines then create a smaller equilateral triangle (shaded area).
What is the ratio of the big equilateral triangle to the small equilateral triangle?
I have seen this problem many years ago.
The inner triangle is one-seventh of the original triangle.
I tried to reproduce the proof and failed abysmally.
I'll try to locate the proof in my vast private library and report back.
[My wife says my library is "half-vast".
. . I think that's what she said.]
I get 1/9, not 1/7.
Assign 3 as length of sides of outer equilateral triangle;
then the three small triangles inside each have one side = 1, angles 20-60-100.
From that, calculate side of inner equilateral triangle:
a = SIN(60) / SIN(20)
b = SIN(100) / SIN(60)
c = SIN(20) / SIN(60)
d = a - b - c (side of inner triangle; will be 1)
Get area of inner and outer triangles:
inner / outer = 1/9
Soroban's memory is right – the answer is 1/7. The flaw in Wilmer's method is that the angles in the small triangle are not 20º and 100º (the line BY in the diagram below trisects the side of the triangle but it does not trisect the angle at B).
Outline proof: apply the cosine rule in triangle BYC to see that if BC = 3 (units) and CY = 1 then BY = √7. The triangles CYX and BYC are similar (equal angles), and the ratio of their sides is 1:√7. So the ratio of their areas is 1:7. But the area of BYC is 1/3 (of the area of the whole triangle), so the area of the small triangles such as CYX is 1/21. It follows that the area of each of the quadrilateral parts of the diagram is 5/21. Thus the total unshaded area is 18/21, and the shaded area (the inner triangle) is 3/21 = 1/7.
I had a plan to solve this.I drew a circumcircle from the centroid of the large triangle and id each of the arcs created starting with the angles at A 9,21,30 (arcs 18 42,60) Each apex angle of the small triangle = 102+18/2=60
For the lengths of the sides of the small triangle I saw that I could do this using coordinate geometry and linear equations but I see now there is a better way.
I believe the OP is missing some critical information in the problem.
There is no way to "solve" the ratio, as the ratio can change depending on how the lines that segment the sides of the large triangle are drawn. Something such as an initial angle or value of two sides must be given. If none of these critical values are given, the ratio can still be figured out, but it would then be a matter of measuring it and not really much of a math problem at all.
If what the OP meant by "solve" is to figure out a theorum/proof for the ratio of the area of the triangles depending on the ratio of the bisected lines of the side of the large triangle, then yes it can and has been "solved" (*I doubt she meant this, but in any case see below).
The original puzzle that I believe the OP wanted to describe is often dubbed "Feynman's triangle" or the "one-seventh triangle" (which is why some of you have believed the ratio to be 1/7th); however, the critical piece of information that the OP excluded is that the lines segment the sides of the large triangle at a ratio of 1:2. Given this knowledge, the problem becomes solvable, as the inner triangle is indeed 1/7th of the larger's, giving a 1:7 ratio. Another important fact that is missed from the OP's triangle, and which makes this problem particularly interesting, is that the triangle need not be equilateral for this ratio to hold true.
According to the wiki entry, Feynman and his friend's came up with a total of four proofs.
One-seventh area triangle - Wikipedia, the free encyclopedia
Here's but one potential proof (*and theorum):
The generalisation to a ratio 1/p is interesting (and the proof for that case is identical to the one that I gave above for the 1/3 case).