A plane is colored with three colors. Prove that there exists a right triangle on this plane, having vertices of the same color. I got stuck with this idea in mind that I could find a shape with all of the vertices connected to each other and all of its angles being 90 degrees. One of such shapes which is not correct is as follows.

But, as you can see, it does not work. What is your suggested manner... shape?
 A: Call the colors $1,2,3$. We can always find two points that have the same color, say $p_1$ and $p_2$ are both colored $1$. Define $p_m$ to be the midpoint of $p_1p_2$. There are two cases:
Case 1: If $p_m$ has color $1$. Then draw three parallel lines $l_1, l_2, l_m$ through the points $p_1,p_2, p_m$, each perpendicular to $p_1p_2$. If any point on $l_1,l_2,l_m$ besides $p_1,p_2,p_m$ has color $1$ we are done. So suppose that every point on $l_1, l_2, l_m$ besides $p_1,p_2,p_m$ has color $2$ or $3$. Choose two points with the same color on $l_1$, call them $q_1,q_2$ and say they are both colored $2$. If we draw lines through $q_1$ and $q_2$ perpendicular to $l_1$, we get $4$ intersection points with lines $l_2, l_m$. If any intersection point has color $2$, we have a monochromatic right triangle with $q_1$ and $q_2$, and if all of the intersetction points are color $3$, they form a monochromatic rectangle, which includes a right triangle. So, every outcome includes a monochromatic right triangle.
Case 2: If $p_m$ has color different from $1$, say $2$. Draw a circle centered at $p_m$ through $p_1$. If any point on the circle other than $p_1$ and $p_2$ has color $1$, it forms a monochromatic right triangle with $p_1$ and $p_2$ and we're done. So suppose every point on the circle except for $p_1$ and $p_2$ has color $2$ or $3$. Take a pair of diameters of the circle at right angles to each other (avoiding the diameter $p_1p_2$). If $3$ or more of the endpoints are color $3$, we are done, so we can assume that at least $2$ of the endpoints have color $2$. If they are antipodal, then with $p_m$, we are back in Case $1$ and if they are not antipodal, they form a monochromatic right triangle with $p_m$.
In every case we find a monochromatic right triangle.
A: Denote any line segment with end points of color x and y as (x,y).
Consider an (r,r). Draw a line perpendicular to it at each end point (l1 and l2). We consider the set A of all line segments between these two lines. Any line of set A can't be (b,b) or (g,g) as then all other points of l1 and l2 will be g or b and we will be done. Any point of l1 and l2 can't be r either. Consider 3 lines from set A. They must be either (b,g) or (g,b). So there are two of a kind(say (b,g)). Extend those line segments(l3 and l4). All other points of l3 and l4 must be r and we are done.
