# Non-congruent angle of an isosceles triangle

Imagine I have two rays R1 and R2 emerging from point A0. Suppose I have N points (X1, X2, X3, . . . , XN) on ray R1, and (Y1, Y2, Y3, . . . , YN) on ray R2 such that X1 and Y1 are the closest to A0 and points are sequentially plotted on these two rays.

Now for every i from 1 to N - 1, we draw a line segment from X[i] to Y[i+1], and from Y[i] to X[i+1]. For i = N, we draw line segment from X[N] to Y[N].

It is given that all these 2N - 1 line segments have equal lengths which itself is equal to |A0X1|, which itself equal to |A0Y1|.

It is obvious that the triangle formed from A0XNYN is isosceles. I want to find the non-congruent angle 'a' or angle XNA0YN. This is given that this angle is < 90 degrees.

I am stuck on it and unable to form a solution.

• Hint: For notational simplicity, let $X$ and $Y$ be points on $R_1$ and $R_2$ that are "farther out" than any $X_i$ or $Y_i$. Now, do a little angle-chasing to determine what each $\angle XX_iY_{i+1}$ (equivalently, $\angle YY_iX_{i+1}$) must be, and notice that $\angle XX_nY_n$ (and $\angle YY_nX_n$) must fit the pattern. – Blue Jan 11 at 10:19
• I have done this already and found the values for each angle but it is dependent upon 'a'. Angle can easily found for any permutation of vertexes but only in terms of 'a', at least for me. I am unable to go ahead. – Anonymous Jan 11 at 10:33
• Got it! I didn't observe that I already had the answer. How foolish? :-D – Anonymous Jan 11 at 11:25
• You should post your solution as an answer, so that we can upvote it (and so that it gets removed from the Unanswered Questions queue). – Blue Jan 11 at 18:32