Geometry problem: a property of the 30-70-80 triangle

Consider a triangle with angles , , . Let be the bisector of the angle . Consider such that . Prove that the triangle is isosceles. 

Weitzenböck’s inequality

Given a triangle , denote by the lengths of the sides opposite to angles . Also denote by the area of the triangle. Weitzenbrock’s inequality states that the following holds:

There are multiple ways of proving this inequality. Perhaps one of the most straightforward ways to approach this is to use the law of sines and trigonometry. We have where is the circumradius. Moreover, is related to by . Thus

Therefore the original inequality is equivalent to

One would be tempted to try the arithmetic geometric mean inequality here, but that cannot work, since the inequality is not homogeneous and are not free, they verify . However, note that the function is concave on therefore, applying Jensen’s inequality we get

As a consequence, using cyclic sums over , we get

where for the last inequality above we applied AM-GM inequality two times. Regrouping we obtain exactly

which is equivalent to the original inequality. Note that with the same ideas we have

showing that we also have

Observe that the equality can hold only if the triangle is equilateral.

This leads to the stronger inequality

Other proofs can be found in the Wikipedia article.

It can be observed that combining the above inequalities we can get the isoperimetric inequality for triangles

where the equality holds if and only if the triangle is equilateral.

Find coefficients of trigonometric polynomial given samples

Suppose the values of some trigonometric polynomial

are known for a set of distinct angles , . 

Question: Recover the coefficients of the trigonometric polynomial that verify when or which best fits the values in the sense of least squares when . 

Answer: Obviously, since there are unknowns, we need at least equations, therefore we restrict ourselves to the case . The equalities produce a set of equations which has at most a solution when , provided the rank of the matrix of the system is . Define the function

This function is a sum of squares, which has zero as a lower bound. The function can be written using norms as where

A straightforward computation shows that the gradient of is

The matrix is invertible, provided all angles are distinct. This is a direct application of the formula of the Vandermonde determinant: using operations on columns you can recover the Vandermonde matrix corresponding to . Therefore, one can always solve the system when and will minimize . In particular, when the minimum will be equal to zero and the coefficients of the trigonometric polynomial verifying will be found. When the best fit, in the sense of least squares, of the values with a trigonometric polynomial will be found.

Below, you can find a Python code which solves the problem in some particular case.

import numpy as np
import matplotlib.pyplot as plt

N = 5 # coeffs
M = 2*N+1 # M>=2N+1 samples

# function to be approximated
def fun(x):
    return np.sin(x+np.sqrt(2))+0.3*np.sin(5*x-np.sqrt(3))+0.1*np.sin(8*x-np.sqrt(7))

thetas =np.linspace(0,2*np.pi,M,endpoint=0)
dthetas =np.linspace(0,2*np.pi,1000,endpoint=1)

# values of the function at sample points
vals = fun(thetas)

A = np.zeros((M,2*N+1))

# construct the matrix of the system
A[:,0] = 1
for i in range(0,N):
    A[:,i+1] = np.cos((i+1)*thetas)
    A[:,N+i+1] = np.sin((i+1)*thetas)

coeffs = np.zeros(2*N+1)


B = (A.T)@A
print(np.shape(B))

# solve the system to find the coefficients
coeffs = np.linalg.solve(B,A.T@vals)

# function computing a trigonometric polynomial
def ptrig(x,c):
    res = np.zeros(np.shape(x))
    res = c[0]
    n = len(c)
    m = (n-1)//2
    for i in range(0,m):
        res = res+c[i+1]*np.cos((i+1)*x)+c[i+m+1]*np.sin((i+1)*x)
    return res

vals2 = ptrig(thetas,coeffs)

# plotting the result
plt.figure()
plt.plot(thetas,vals,'.b',label="Sample values")
plt.plot(dthetas,fun(dthetas),'g',label="Original function")
plt.plot(dthetas,ptrig(dthetas,coeffs),'r',label="Fitted trigonometric polynomial")
plt.legend()
plt.savefig("TrigPoly.png",dpi=100,bbox_inches='tight')
plt.show()

For the parameters chosen above the program outputs the following result. You can play with the input parameters to observe the changes.

Putnam 2019 – Problem B2

B2. For all , let

Determine

Putnam 2019 A2

A2. In the triangle , let be the centroid, and let be the center of the inscribed circle. Let and be the angles at the vertices and , respectively. Suppose that the segment is parallel to and that . Find .

Putnam 2018 – Problem B4

B4. Given a real number , we define a sequence by , and for . Prove that if for some , then the sequence is periodic.

Putnam 2018, Problem B4

Solution: The path to the solution for this problem is really interesting. The first question that pops in mind when seeing the problem is: How do I prove that a sequence is periodic?. It’s not obvious from the given recurrence relation that this should happen, and what’s with the condition ?