MaxWong

When x = y = 0, \((f(0))^2 = 2f(0)\).

This implies f(0) = 0 or f(0) = 2.

Case 1: f(0) = 2.

When x = 0, 2 f(y) = f(y) + 2

f(y) = 2 for all real numbers y.

The function f(x) = 2 is a solution.

Case 2: f(0) = 0.

When x = 0, 0 = f(y) for all real numbers y.

The function f(x) = 0 is a solution.

Therefore, the possible functions are the constant functions f(x) = 0 and f(x) = 2.

We can choose 3 distinct points to form a triangle, but there are some cases that the 3 points do not form a triangle (for example the 3 points are all on the same diagonal/row/column). We need to subtract those cases from the number of ways to choose 3 distinct points from 9.

We can count the number of cases of collinear points by hand. 

All on same row: 3 cases

All on same column: 3 cases

All on same diagonal: 2 cases

Total: 8 cases that do not form a triangle. We subtract that from 9C3:

\(\text{Number of ways} = \binom{9}3 - 8 = 76\)

We first move all terms to the left-hand side.

\(x^2 + 10x - 4x - 9= 0 \\ x^2 + 6x - 9= 0\)

We want to make (x + r)^2 on the left-hand side, and we know that (x + 3)^2 = x^2 + 6x + 9. What if we add 18 to both sides?

\(x^2 + 6x + 9 = 18\)

Now just write the left-hand side as (x + 3)^2 and there's the answer.

I used this symbol for the non-real root for a reason. It is called the "cube roots of unity". Generally, any root of unity is commonly denoted by \(\omega\). 

WIki page: https://en.wikipedia.org/wiki/Root_of_unity

Suppose that the non-real root is \(\omega\). Obviously, since it is non-real, it cannot be equal to 1.

Since its cube is 1, \(\omega^3 = 1\), which means \(\omega^3 - 1 = 0\). If we factorize the left-hand side of the equation, we have \((\omega - 1)(\omega^2 + \omega + 1) = 0\). Since \(\omega\) is not 1, we can cross out \((\omega - 1)\), so we're left with \(\omega^2 + \omega + 1 = 0\).

With the derivation above, it turns out that if \(\omega\) is a non-real root whose cube is 1, then \(x = \omega\) must be a root of the equation \(x^2 + x + 1 = 0\). Therefore, a = b = 1 by comparing coefficients.

This one is similar to this problem. Please read the solution there and try to imitate it.

https://web2.0calc.com/questions/geometry_65851

Hint: t^2 - 12t - 36 = (t - 6)^2 - 72. If you want t^2 - 12t - 36 to be the smallest, then you want (t - 6)^2 to be the smallest, but (t - 6)^2 is always >= 0. How can we make (t - 6)^2 = 0, so that the expression t^2 - 12t - 36 is minimized?

We can use Law of Cosines twice.

First, we use it on triangle ABC to find the cosine of angle B.

\(\cos \angle B = \dfrac{3^2 + 7^2 - 5^2}{2(3)(7)}\\ \cos \angle B = \dfrac{11}{14}\)

Then, we use it again, but this time on triangle ABD.

\(AD^2 = AB^2 + BD^2 - 2(AB)(BD) \cos \angle B\\ AD^2 = 3^2 + \left(\dfrac72\right)^2 - 2(3)\left(\dfrac72\right)\left(\dfrac{11}{14}\right)\\ AD^2 = \dfrac{19}4\\ AD = \dfrac{\sqrt{19}}2\text{ cm}\)

It is given that angle OQA = 90 degrees, which means \((\text{slope of }OQ) (\text{slope of }QA) = -1\).

Let Q = (x, x + 8). Then \(\dfrac{(x + 8) - 0}{x - 0}\cdot \dfrac{(x + 8) - (-10)}{x - 10} = -1\)

\((x + 8)(x + 18) + x(x - 10) = 0\)

If you expand and solve the equation, you will find that this equation has no real solutions.

It turns out that the line y = x + 8 is too far away from O and A, so there are no such point on Q such that angle OQA = 90 degrees. Even the point on y = x + 8 closest to the midpoint of OA gives an acute angle OQA.

We can make use of the identity \(\cos \theta = \sin(90^\circ - \theta)\).

\(\sin 3\theta = \cos 6\theta\\ \sin 3\theta = \sin(90^\circ - 6\theta)\)

To get the smallest positive angle theta, we remove the sin from both sides (note that it does not always work, but if we assume that 0 degrees < 3 theta < 90 degrees and 0 degrees < 90 degrees - 6 theta < 90 degrees, then it works.)

\(3\theta = 90^\circ -6\theta\\ 9\theta = 90^\circ\\ \theta = \boxed{10^\circ}\)