MaxWong

\(\begin{cases} x = y^4\\ x + y^2 = 1 \end{cases}\)

Substituting,

\(y^4 + y^2 - 1= 0\\ y^2 = \dfrac{-1 \pm \sqrt 5}2\text{(reject negative root)}\\ y^2 = \dfrac{\sqrt 5 - 1}2\\ y = \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\)

Since \(x + y^2 = 1\),

\(x = 1 - \dfrac{\sqrt 5 - 1}2 = \dfrac{3 -\sqrt 5}2\)

So the intersection points are \(\left(\dfrac{3 - \sqrt 5}2, \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\right)\).

The distance is \(2 \sqrt {\dfrac{\sqrt 5 - 1}2} = \sqrt{2(\sqrt 5 - 1)} = \sqrt{-2 + 2 \sqrt 5}\).

Then \((u,v) = (-2, 2)\).

\(\begin{cases} x^2 + 4y^2 = 4\\ x^2 - m(y + 2)^2 = 1 \end{cases}\)

Substituting, 

\(4 - 4y^2 - m(y + 2)^2 = 1\\ m(y + 2)^2 + 4y^2 - 3 = 0\\ (m + 4)y^2 + 4my + 4m - 3 = 0\)

Take \(\Delta = 0\), we have 

\((4m)^2 - 4(m + 4)(4m - 3) = 0\\ 16m^2 - 4(4m^2 +13m -12) = 0\\ -52m+48 = 0\\ m = \dfrac{12}{13}\)

Suppose the legs are a and b. Then by the definition of cool trianlges,

\(\dfrac{ab}2 = 2(a + b)\\ ab = 4a + 4b\\ ab - 4a - 4b + 16 = 16\\ (a - 4)(b - 4) = 16\)

Assume without loss of generality that \(a \leq b\). Then

\(\begin{cases} a-4=1\\ b-4=16 \end{cases}\text{ or }\begin{cases} a-4=2\\ b-4=8 \end{cases}\text{ or }\begin{cases} a-4=4\\ b-4=4 \end{cases} \)

This gives all three cool triangles, the one with legs 5 and 20, the one with legs 6 and 12, and the one with both legs equal to 8.

The sum of all possible areas is \(\dfrac{5 \times 20 + 6 \times 12 + 8 \times 8}2 = 118\).

Let AB = b, and AC = c. Moreover, let \([ABX] = A_1\) and \([ACX] = A_2\), where \([\cdot]\) denote "area of".

By Stewart's theorem, we have 

\((13 + 5)(13^2 + 13(5)) = 5b^2 + 13c^2\\ 5b^2 + 13c^2 = 4212 \)

By basic geometry, we have \(\dfrac{A_1}{A_2} = \dfrac{13}5\). 

Recall that the circumradius of a triangle with sides a, b, c and area A is \(\dfrac{abc}{4A}\). Then, since the circumradius of triangles ABX and ACX are the same, we have 

\(\dfrac{13^2 b}{4A_1} = \dfrac{5(13)c}{4A_2}\\ \dfrac{A_1}{A_2} = \dfrac{13b}{5c}\\ \dfrac{13}5 = \dfrac{13b}{5c}\\ b = c\).

Hence, 

\(18b^2 = 18c^2 = 4212\\ b = c = 3 \sqrt{26}\)

We calculate the half perimeter of triangle ABC to be \(\dfrac{18 + b + c}2 = 3 \sqrt{26} + 9\). Then, by Heron's formula, the area required is:

\(\sqrt{(3 \sqrt{26} + 9)(3\sqrt{26} - 9)(9^2)} = 27\sqrt{17}\). 

\(m_{BC } = \dfrac{2 - (-8)}{-3 - 1} = -\dfrac52\\ m_{\text{altitude}} = \dfrac25\\ \dfrac{y - 3}{0 - 2} = \dfrac25\\ y - 3 = -\dfrac45\\ y = \dfrac{11}5\)

\(1000 \left(-\dfrac89\right)^{n-1} > 1\)

To be greater than 1, it must be greater than 0, so we only count the odd numbered terms, i.e. 1st term, 3rd term, etc.

After some point, even if it is an odd numbered term, it won't be greater than 1 anymore.

Since the term is an odd numbered term as we deduced, we can safely let n = 2k + 1 for k >= 0, k is an integer.

\(1000 \left(\dfrac{64}{81}\right)^k > 1\\ k < \log_{64/81}{\dfrac1{1000}}\\ k < 29.3\)

Then any nonnegative integer k with k < 29 will correspond to a positive term in the original sequence which is greater than 1.

Hence, the answer is 29.

We will tackle the general case first. What is the sum of all positive integers less than 1000 ending in a digit n?

The numbers must be of the form \(10m + n\) where \(0 \leq m \leq 99\).Therefore, the answer will be \(\displaystyle \sum_{m = 0}^{99} (10m + n) = 100n + 49500\).

Hence, the answer to your original question will be \(4(49500) + 100(3 + 4 + 6 + 9) = 200200\).

Simply divide by the amount you have overcounted using 7!.

\(\text{Answer} = \dfrac{7!}{3!2!2!} = 210\)