Rom

legs of a triangle a, b, c

the triangle inequality states

\(|a|+|b| >|c| \\ \text{with strict inequality else we'd just have a line segment, not a triangle.}\)

\(|c| < |39|+|15| = 54 \Rightarrow c < 54 \\ \text{we also must have }\\ |c|+|15| > 39 \Rightarrow c > 24 \text{ thus}\\ 24 < c < 54 \)

\(\text{If I understand what you're saying...}\\ 4\sqrt{abc}=5\\ 4\sqrt{abcd}=3\sqrt{10}\)

\(\dfrac{4\sqrt{abcd}}{4\sqrt{abd}}=\sqrt{d} = \dfrac{3\sqrt{10}}{5}\\ d = \left(\dfrac{3\sqrt{10}}{5}\right)^2 = \dfrac{9\cdot 10}{25} = \dfrac{18}{5}\)

\(\text{assuming a 365 day year}\\ nblink=3\dfrac{blink}{min}\cdot 60\dfrac{min}{hr}\cdot 14\dfrac{hr}{day}\cdot 365 \dfrac{day}{yr} = \\ 3\cdot 60 \cdot 14 \cdot 365 =919800 \dfrac{blink}{yr} \\ \text{so no, not quite a million blinks per year}\)

I think the easiest way to do this is to parameterize the circle via a single variable and solve the minimization problem.

\(\text{let }p \text{ be a point on the circle given by }(x-12)^2 + (y+5)^2 = 9 \\ \text{then p can be described as }\\ p=(3 \cos(\theta)+12,~3\sin(\theta)-5)\)

\(\text{The squared distance of this point to the origin is }\\ d^2 = (3 \cos(\theta)+12)^2 + (3\sin(\theta)-5)^2 =\\ 9 \sin ^2(\theta )-30 \sin (\theta )+9 \cos ^2(\theta )+72 \cos (\theta )+169 = \\ -30 \sin (\theta )+72 \cos (\theta )+178 = \\ 78 \cos(\theta+\arctan(5/12))+178\)

\(d^2 \text{will clearly be minimized when }\\ \theta + \arctan(5/12) = \pi\\ \theta = \pi - \arctan(5/12)\\ \text{at this angle }d^2 = -78 + 178 = 100\\ d=10\)

The key here is to find x such that the determinant of F is zero.

\(det(F) = 2\cdot 3 - (-4)\cdot x = 6+4x \\ \text{solving }det(F)=0 \text{ we get} \\ 6+4x = 0 \\ x = -\dfrac 3 2\)

well let's multiply them

\(\begin{pmatrix}4 &3\\1&1\end{pmatrix} \begin{pmatrix}1 &-3\\-1&4\end{pmatrix} =\begin{pmatrix}1&0\\0&1\end{pmatrix}\)

so yes C and E are multiplicative inverses

There are a total of \(\dbinom{6}{2}=15\) possible angles

But it turns out that many of the angles are congruent.

The 15 angles are made up of 1 + 2 + 3 + 4 + 5 copies of 5 different angles.

There are only 5 different possible angles

\(\text{This is known as finding the partial fraction expansion}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{x+y}{(x-2)(x+1)} = \dfrac{A}{x-2}+\dfrac{B}{x-1}\)

\(\dfrac{x+7}{(x-2)(x+1)} = \dfrac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \forall x : x \neq 2,~-1 \\ x+7 = A(x+1) + B(x-2) \\ x = (A+B)x \Rightarrow \\ 1 = A+B\\ 7 = A -2B \)

\(B = 1-A\\ 7 = A - 2(1-A) = 3A-2\\ 3A = 9\\ A=3\\ B = 1-3 = -2\\ \dfrac{x+7}{(x-2)(x+1)}=\dfrac{3}{x-2}-\dfrac{2}{x+1}\)

\(\text{let }n_c \text{ be the number of cups of lemonade sold and }\\n_b \text{ be the number of lemon bars sold.}\)

\(\text{we have 3 constraints }\\ 2\cdot n_c + 1.5 \cdot n_b \geq 500 \\ 0.25 n_c + 0.2 n_b \leq 100\\ n_c \geq 150\)

Oh oops, I missed that they called n_c = x and n_b = y, oh well you'll figure it out.

I don't see any solution method any better than just plugging in the points and checking.

so every pair but the first listed is acceptable.

The theorem being used here is that

\(gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1\)

in this case \(n=1001,~m=1012\)

\(1001 = 7 \cdot 11 \cdot 13 \\ 1012 = 2 \cdot 11 \cdot 23 \\ \text{so clearly }gcd(1001,~1012)=11\\ 2^{11}-1 = 2047 \text{ as guest correctly computed}\)

The proof of this theorem can be found online