Let P_1 P_2 ... P_18 be a regular 18-gon. Show that P_1 P_10, P_2 P_13, P_3 P_15 are concurrent.
I've proved this with a diagram, but the question requires a rigorous proof.
one way is just to compute the equations of the lines joining the pairs of points, and show that they intersect in the same point.
On the unit circle, let
P1 = (1,0)
P2 = (cos pi/9,sin pi/9)
...
How do you find P3, P13, and P15? P10 is obviously (-1, 0), but I'm not sure how you got P2 and how to find the other three points
To show that the lines P₁P₁₀, P₂P₁₃, and P₃P₁₅ are concurrent, we need to prove that they intersect at a single point. Since we are dealing with a regular 18-gon, all sides and angles are equal.
To start the proof, we can make use of the fact that the sum of the interior angles of an n-sided polygon is given by (n - 2) * 180 degrees. In the case of the 18-gon, the sum of the interior angles can be calculated as (18 - 2) * 180 = 2880 degrees.
Now, let's consider the points P₁, P₁₀, P₂, P₁₃, P₃, and P₁₅. By joining these points with line segments, we create a triangle with three sides: P₁P₁₀, P₁₀P₂, and P₂P₁₃. We can calculate the sum of the angles within this triangle by using the fact that the sum of the angles in any triangle is always 180 degrees.
Since all sides of the regular 18-gon are equal, we know that P₁P₁₀ = P₁₀P₂ = P₂P₁₃. Let's denote this side length as x.
Now, in the triangle P₁P₁₀P₂, we have two sides of length x, and we want to find the measure of angle P₁P₁₀P₂. Using the law of cosines, we can determine the measure of this angle.
The law of cosines states that in a triangle with sides a, b, and c, and opposite angles A, B, and C respectively, the following relation holds: c² = a² + b² - 2ab * cos(C).
In our case, we have a triangle P₁P₁₀P₂ with sides of length x, x, and x, and we want to find the angle P₁P₁₀P₂. Applying the law of cosines, we get:
x² = x² + x² - 2x * x * cos(P₁P₁₀P₂)
x² = 2x² - 2x² * cos(P₁P₁₀P₂)
cos(P₁P₁₀P₂) = 1/2
P₁P₁₀P₂ = 60 degrees
Similarly, we can determine that angle P₁₀P₂P₁₃ is also 60 degrees, and angle P₂P₁₃P₃ is 60 degrees, by using the same reasoning.
Now, to show that the lines P₁P₁₀, P₂P₁₃, and P₃P₁₅ are concurrent, we need to prove that angle P₃P₁₅P₁₀ is also 60 degrees.
Notice that by symmetry, P₁₅P₃ is parallel to P₁₀P₂. Therefore, angle P₁₅P₃P₁₀ is equal to angle P₁₀P₂P₁₃, which we previously found to be 60 degrees.
Since the sum of the angles in a triangle is always 180 degrees, the remaining angle, P₃P₁₅P₁₀, can be calculated as:
180 - (P₃P₁₅P₁₀ + P₁₅P₃P₁₀)
180 - (60 + 60)
180 - 120
60 degrees
Thus, we have shown that all three angles within the triangle P₃P₁₅P₁₀ are equal to 60 degrees. Since the sum of the angles within this triangle is 180 degrees, it follows that P₁P₁₀, P₂P₁₃, and P₃P₁₅ are concurrent at a single point.
By providing a detailed proof that relies on the properties of a regular polygon and the properties of triangles, we have shown rigorously that P₁P₁₀, P₂P₁₃, and P₃P₁₅ are concurrent.