# How many isosceles triangles can be made in the x-y plane that satisfy all of the following conditions: a. Integer coordinates, b. Area = 9, c. A vertex at the origin?

They are $36$ (two triangles that differ only by vertex labeling are considered to be the same triangle, so they count as one).

#### Explanation:

PREMISE:
I hope that someone will find the time to check this and maybe come up with a smarter (and shorter) way to solve the problem. In the following, I'm going to solve it in a "semi-bruteforce" way.

CONDITIONS:

Let's try to translate the conditions into mathematical language.

Condition [c] says that $O = \left(0 , 0\right)$ is a vertex of the triangle. We have a triangle $A B O$ that has to be isosceles, so one of the following has to hold:

• $\hat{A} = \hat{O}$ and $A B = B O$
• $\hat{B} = \hat{O}$ and $A B = A O$
• $\hat{A} = \hat{B}$ and $A O = B O$

Condition [a] says that ${x}_{A} , {x}_{B} , {y}_{A} , {y}_{B} \in \mathbb{Z}$.

Condition [b] can be translated into a condition on the cross product between $\vec{O A}$ and $\vec{O B}$: $S = \frac{1}{2} | | \vec{O A} \times \vec{O B} | | = \frac{1}{2} | | \vec{O A} | | \setminus | | \vec{O B} | | \setminus | \sin \hat{O} | = 9$ (where $| | \cdot | |$ is the magnitude of the vector and $| \cdot |$ is the absolute value). So $\sin \hat{O} = \pm \frac{18}{A O \setminus B O}$ and if $A O \setminus B O \ge 18$ (otherwise we get unacceptable cases) we conclude that $\hat{O} = \arcsin \left(\pm \frac{18}{A O \setminus B O}\right)$.

So we translated the problem into mathematical language and we got limitation : $A O \setminus B O \setminus \ge 18$.

FIRST MANIPULATIONS:

By the Pythagorean Theorem, the following identities hold:

• $A {O}^{2} = {x}_{A}^{2} + {y}_{A}^{2}$
• $B {O}^{2} = {x}_{B}^{2} + {y}_{B}^{2}$
• $A {B}^{2} = {\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2}$

Note: by condition [a] we have that $A {O}^{2} , B {O}^{2} , A {B}^{2} \in \mathbb{Z}$. This doesn't necessarily imply that $A O , B O , A B \in \mathbb{Z}$!

The last identity can be rewritten as follows (also known as Law of Cosines):
$A {B}^{2} = A {O}^{2} + B {O}^{2} - 2 \left({x}_{A} {x}_{B} + {y}_{A} {y}_{B}\right) = A {O}^{2} + B {O}^{2} - 2 \left(\vec{O A} \cdot \vec{O B}\right) = A {O}^{2} + B {O}^{2} - 2 \setminus A O \setminus B O \cos \hat{O}$

Substituting $\hat{O}$ given by condition [b] and using the fact that $\cos \left(\arcsin \left(x\right)\right) = \sqrt{1 - {x}^{2}}$ if $- 1 \le x \le 1$, we get
$A {B}^{2} = A {O}^{2} + B {O}^{2} - 2 \setminus A O \setminus B O \cos \left[\arcsin \left(\pm \frac{18}{A O \setminus B O}\right)\right] = A {O}^{2} + B {O}^{2} - 2 \setminus A O \setminus B O \sqrt{1 - {18}^{2} / \left(A {O}^{2} \setminus B {O}^{2}\right)} = A {O}^{2} + B {O}^{2} - 2 \sqrt{A {O}^{2} \setminus B {O}^{2} - {18}^{2}}$
This can be rewritten as the identity 
$A {O}^{2} + B {O}^{2} - A {B}^{2} = 2 \sqrt{A {O}^{2} \setminus B {O}^{2} - {18}^{2}}$

FIRST CASE:

We have three cases to analyze. Let's start with the first case: $\hat{A} = \hat{O}$ and $A B = B O$.

The identity  becomes $A {O}^{2} = 2 \sqrt{A {O}^{2} \setminus B {O}^{2} - {18}^{2}}$, so we can square the two (positive) sides and get
$A {O}^{4} = 4 \setminus A {O}^{2} \setminus B {O}^{2} - 4 \cdot {18}^{2}$
$4 \setminus A {O}^{2} \setminus B {O}^{2} - A {O}^{4} = {36}^{2}$
$A {O}^{2} \left(4 B {O}^{2} - A {O}^{2}\right) = {2}^{4} \cdot {3}^{4}$

By condition [a] we know that $A {O}^{2} = {x}_{A}^{2} + {y}_{A}^{2} \in \mathbb{Z}$ and $4 B {O}^{2} - A {O}^{2} = 4 {x}_{B}^{2} + 4 {y}_{B}^{2} - {x}_{A}^{2} - {y}_{A}^{2} \in \mathbb{Z}$. This means that both factors have to be integer divisors of ${2}^{4} \cdot {3}^{4}$, their product has to be ${2}^{4} \cdot {3}^{4}$ and they have equal sign. $A {O}^{2}$ is positive, so $4 B {O}^{2} - A {O}^{2} \ge 0$ i.e. $A {O}^{2} \le 4 B {O}^{2}$.
The solutions to the equation obtained above are among the following candidates: $A {O}^{2} = {2}^{k} \cdot {3}^{h}$ and $4 B {O}^{2} - A {O}^{2} = {2}^{4 - k} \cdot {3}^{4 - h}$ for $h , k = 0 , 1 , 2 , 3 , 4$ (25 cases).

By condition [a] $B {O}^{2} = {x}_{B}^{2} + {y}_{B}^{2} \in \mathbb{Z}$, then $4 B {O}^{2} = {2}^{4 - k} \cdot {3}^{4 - h} + A {O}^{2} = {2}^{4 - k} \cdot {3}^{4 - h} + {2}^{k} \cdot {3}^{h}$ has to be divisible by $4$.

• If $k = 0$, then $4 B {O}^{2} = {3}^{4 - h} + {2}^{4} \cdot {3}^{h}$ is an addition between an odd number and an even number, so the sum has to be odd; this leads to a contradiction ($4 B {O}^{2}$ is divisible by $4$ and therefore even), so the solutions for $k = 0$ have to be rejected.
• If $k = 4$, we follow an analogous argument to reject solutions of this form.
• If $k = 1$, then $4 B {O}^{2} = 2 \cdot {3}^{4 - h} + {2}^{3} \cdot {3}^{h}$. It follows that $2 B {O}^{2} = {3}^{4 - h} + {2}^{2} \cdot {3}^{h}$ is an addition between an odd and an even number, so the sum $2 B {O}^{2}$ is not even and this is a contradiction. Then, solutions for $k = 1$ have to be rejected.

• If $k = 3$, we follow an analogous argument to reject solutions of this form.

• If $k = 2$, then $A {O}^{2} = {2}^{2} \cdot {3}^{h}$ and $4 B {O}^{2} = {2}^{2} \cdot {3}^{4 - h} + {2}^{2} \cdot {3}^{h}$ i.e. $B {O}^{2} = {3}^{4 - h} + {3}^{h}$.

So the only acceptable candidates are the ones for which $k = 2$. Two conditions should hold:

• $A O \setminus B O \ge 18 \iff A {O}^{2} \setminus B {O}^{2} \ge {18}^{2}$ and substituting we get ${3}^{4} + {3}^{2 h} \ge {3}^{4}$ and this is satisfied for every $h = 0 , 1 , 2 , 3 , 4$.
• $A {O}^{2} \le 4 B {O}^{2} \iff {2}^{2} \cdot {3}^{h} \le {2}^{2} \cdot {3}^{4 - h} + {2}^{2} \cdot {3}^{h}$ i.e. ${3}^{h} \le {3}^{4 - h} + {3}^{h}$, which is true for every $h = 0 , 1 , 2 , 3 , 4$.

So we are left with the following solution candidates ($k = 2$):

• $h = 0$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {2}^{2} \cdot {3}^{0} = 4$ and $4 = {0}^{2} + {2}^{2}$, so we have $\left({x}_{A} , {y}_{A}\right) = \left(0 , \pm 2\right)$ or $\left({x}_{A} , {y}_{A}\right) = \left(\pm 2 , 0\right)$; moreover ${x}_{B}^{2} + {y}_{B}^{2} = B {O}^{2} = {3}^{4} + {3}^{0} = 82$ and $82 = {1}^{2} + {9}^{2}$, so we have $\left({x}_{B} , {y}_{B}\right) = \left(\pm 1 , \pm 9\right)$ or $\left({x}_{B} , {y}_{B}\right) = \left(\pm 9 , \pm 1\right)$.
• $h = 1$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {2}^{2} \cdot {3}^{1} = 12$ and $12$ can't be written as a sum of two squared integers.
• $h = 2$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {2}^{2} \cdot {3}^{2} = 36$ and $36 = {0}^{2} + {6}^{2}$, so we have $\left({x}_{A} , {y}_{A}\right) = \left(0 , \pm 6\right)$ or $\left({x}_{A} , {y}_{A}\right) = \left(\pm 6 , 0\right)$; moreover ${x}_{B}^{2} + {y}_{B}^{2} = B {O}^{2} = {3}^{2} + {3}^{2} = 18$ and $18 = {3}^{2} + {3}^{2}$, so we have $\left({x}_{B} , {y}_{B}\right) = \left(\pm 3 , \pm 3\right)$.
• $h = 3$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {2}^{2} \cdot {3}^{3} = 108$ and $108$ can't be written as a sum of two squared integers.
• $h = 4$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {2}^{2} \cdot {3}^{4} = 324$ and $324 = {0}^{2} + {18}^{2}$, so we have $\left({x}_{A} , {y}_{A}\right) = \left(0 , \pm 18\right)$ or $\left({x}_{A} , {y}_{A}\right) = \left(\pm 18 , 0\right)$; moreover ${x}_{B}^{2} + {y}_{B}^{2} = B {O}^{2} = {3}^{0} + {3}^{4} = 82$ and $82 = {1}^{2} + {9}^{2}$, so we have $\left({x}_{B} , {y}_{B}\right) = \left(\pm 1 , \pm 9\right)$ or $\left({x}_{B} , {y}_{B}\right) = \left(\pm 9 , \pm 1\right)$.

So, considering that in this first case $A B = B O$, we get the following solutions:

1. $A = \left(0 , 2\right)$ and $B = \left(9 , 1\right)$
2. $A = \left(0 , 2\right)$ and $B = \left(- 9 , 1\right)$
3. $A = \left(0 , - 2\right)$ and $B = \left(9 , - 1\right)$
4. $A = \left(0 , - 2\right)$ and $B = \left(- 9 , - 1\right)$
5. $A = \left(2 , 0\right)$ and $B = \left(1 , 9\right)$
6. $A = \left(2 , 0\right)$ and $B = \left(1 , - 9\right)$
7. $A = \left(- 2 , 0\right)$ and $B = \left(- 1 , 9\right)$
8. $A = \left(- 2 , 0\right)$ and $B = \left(- 1 , - 9\right)$
9. $A = \left(0 , 6\right)$ and $B = \left(3 , 3\right)$
10. $A = \left(0 , 6\right)$ and $B = \left(- 3 , 3\right)$
11. $A = \left(0 , - 6\right)$ and $B = \left(3 , - 3\right)$
12. $A = \left(0 , - 6\right)$ and $B = \left(- 3 , - 3\right)$
13. $A = \left(6 , 0\right)$ and $B = \left(3 , 3\right)$
14. $A = \left(6 , 0\right)$ and $B = \left(3 , - 3\right)$
15. $A = \left(- 6 , 0\right)$ and $B = \left(- 3 , 3\right)$
16. $A = \left(- 6 , 0\right)$ and $B = \left(- 3 , - 3\right)$
17. $A = \left(0 , 18\right)$ and $B = \left(1 , 9\right)$
18. $A = \left(0 , 18\right)$ and $B = \left(- 1 , 9\right)$
19. $A = \left(0 , - 18\right)$ and $B = \left(1 , - 9\right)$
20. $A = \left(0 , - 18\right)$ and $B = \left(- 1 , - 9\right)$
21. $A = \left(18 , 0\right)$ and $B = \left(9 , 1\right)$
22. $A = \left(18 , 0\right)$ and $B = \left(9 , - 1\right)$
23. $A = \left(- 18 , 0\right)$ and $B = \left(- 9 , 1\right)$
24. $A = \left(- 18 , 0\right)$ and $B = \left(- 9 , - 1\right)$

SECOND CASE:

$\hat{B} = \hat{O}$ and $A B = B O$.

We can act in the same way we did in the first case. This produces exactly the same solutions listed in the first case (with $A$ playing the role of $B$ and vice versa), so no new triangle can be found in this manner.

THIRD CASE:

$\hat{A} = \hat{B}$ and $A O = B O$.

The identity  becomes
$2 A {O}^{2} - A {B}^{2} = 2 \sqrt{A {O}^{4} - {18}^{2}}$, so we can square both sides under the condition that $2 A {O}^{2} \ge A {B}^{2}$ (otherwise solutions are not acceptable since they're not real) and get
${\left(2 A {O}^{2} - A {B}^{2}\right)}^{2} = 4 \left(A {O}^{4} - {18}^{2}\right)$
$4 A {O}^{4} - 4 A {O}^{2} A {B}^{2} + A {B}^{4} = 4 A {O}^{4} - {36}^{2}$
$A {B}^{2} \left(4 A {O}^{2} - A {B}^{2}\right) = {36}^{2}$
$\left(4 A {O}^{2} - A {B}^{2}\right) A {B}^{2} = {2}^{4} \cdot {3}^{4}$

$A {B}^{2} = {2}^{k} \cdot {3}^{h}$ and $4 A {O}^{2} - A {B}^{2} = {2}^{4 - k} \cdot {3}^{4 - h}$, so $4 A {O}^{2} = {2}^{k} \cdot {3}^{h} + {2}^{4 - k} \cdot {3}^{4 - h}$ for $h , k = 0 , 1 , 2 , 3 , 4$. Note that $4 A {O}^{2} \ge A {B}^{2}$ holds because we already requested that $2 A {O}^{2} \ge A {B}^{2}$.
Since $A {O}^{2} \in \mathbb{Z}$, by the same argument used in the first case for $B O$ in place of $A B$, we conclude that $4 A {O}^{2}$ is divisible by $4$ if and only if $k = 2$. So $A {B}^{2} = 4 \cdot {3}^{h}$, $4 A {O}^{2} - A {B}^{2} = 4 \cdot {3}^{4 - h}$ and $A {O}^{2} = {3}^{h} + {3}^{4 - h}$.

We have to check two conditions:

• $A {O}^{2} \ge 18 \iff {3}^{h} + {3}^{4 - h} \ge 18$ and this is satisfied for each $h = 0 , 1 , 2 , 3 , 4$.
• $2 A {O}^{2} \ge A {B}^{2} \iff 2 \cdot {3}^{h} + 2 \cdot {3}^{4 - h} \ge 4 \cdot {3}^{h}$ i.e. ${3}^{4 - h} \ge {3}^{h}$ that is satisfied only if $h = 0 , 1 , 2$.

So we are left with the following solution candidates:

• If $h = 0$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {3}^{0} + {3}^{4} = 82$ and $82 = {1}^{2} + {9}^{2}$, so we have $\left({x}_{A} , {y}_{A}\right) = \left(\pm 1 , \pm 9\right)$ or $\left({x}_{A} , {y}_{A}\right) = \left(\pm 9 , \pm 1\right)$; moreover ${\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2} = A {B}^{2} = 4 \cdot {3}^{0} = 4$ and $4 = 0 + {2}^{2}$, so we have $\left({x}_{A} - {x}_{B} , {y}_{A} - {y}_{B}\right) = \left(0 , \pm 2\right)$ or $\left({x}_{A} - {x}_{B} , {y}_{A} - {y}_{B}\right) = \left(\pm 2 , 0\right)$ (from these we will easily derive the possible values for ${x}_{B}$ and ${y}_{B}$, given a combination of ${x}_{A}$ and ${y}_{A}$).

• if $h = 1$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {3}^{1} + {3}^{3} = 30$ and $30$ can't be written as a sum of two squared integers.

• if $h = 2$, then ${x}_{A}^{2} + {y}_{A}^{2} = A {O}^{2} = {3}^{2} + {3}^{2} = 18$ and $18 = {3}^{3} + {3}^{2}$, so we have $\left({x}_{A} , {y}_{A}\right) = \left(\pm 3 , \pm 3\right)$; moreover ${\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2} = A {B}^{2} = 4 \cdot {3}^{2} = 36$ and $36 = 0 + {6}^{2}$, so we have $\left({x}_{A} - {x}_{B} , {y}_{A} - {y}_{B}\right) = \left(0 , \pm 6\right)$ or $\left({x}_{A} - {x}_{B} , {y}_{A} - {y}_{B}\right) = \left(\pm 6 , 0\right)$.

So, considering that in this third case $A O = B O$, we get the following solutions:

1. $A = \left(1 , 9\right)$ and $B = \left(1 , - 9\right)$
2. $A = \left(1 , 9\right)$ and $B = \left(- 1 , 9\right)$
3. $A = \left(- 1 , 9\right)$ and $B = \left(- 1 , - 9\right)$
4. $A = \left(- 1 , - 9\right)$ and $B = \left(1 , - 9\right)$
5. $A = \left(9 , 1\right)$ and $B = \left(9 , - 1\right)$
6. $A = \left(9 , 1\right)$ and $B = \left(- 9 , 1\right)$
7. $A = \left(9 , - 1\right)$ and $B = \left(- 9 , - 1\right)$
8. $A = \left(- 9 , - 1\right)$ and $B = \left(- 9 , 1\right)$
9. $A = \left(3 , 3\right)$ and $B = \left(3 , - 3\right)$
10. $A = \left(3 , 3\right)$ and $B = \left(- 3 , 3\right)$
11. $A = \left(- 3 , 3\right)$ and $B = \left(- 3 , - 3\right)$
12. $A = \left(- 3 , - 3\right)$ and $B = \left(3 , - 3\right)$

CONCLUSION

We proved that there are $24 + 12 = 36$ different isosceles triangles $A B O$ that satisfy the three conditions given. Note that two triangles that differ only by vertex labeling ($A$ and $B$ roles inverted) are considered to be the same and together count as one.

Nov 21, 2015

Maybe it's easier to check all possibilities of four integers in $\left[- 18 , 18\right] \cap m a t h \boldsymbol{Z}$.

#### Explanation:

1) ${z}_{1} = - 18 , {z}_{2} = 0 , {z}_{3} = - 9 , {z}_{4} = - 1$
2) ${z}_{1} = - 18 , {z}_{2} = 0 , {z}_{3} = - 9 , {z}_{4} = 1$
3) ${z}_{1} = - 9 , {z}_{2} = - 1 , {z}_{3} = - 18 , {z}_{4} = 0$
4) ${z}_{1} = - 9 , {z}_{2} = - 1 , {z}_{3} = - 9 , {z}_{4} = 1$
5) ${z}_{1} = - 9 , {z}_{2} = - 1 , {z}_{3} = 0 , {z}_{4} = - 2$
6) ${z}_{1} = - 9 , {z}_{2} = - 1 , {z}_{3} = 9 , {z}_{4} = - 1$
7) ${z}_{1} = - 9 , {z}_{2} = 1 , {z}_{3} = - 18 , {z}_{4} = 0$
8) ${z}_{1} = - 9 , {z}_{2} = 1 , {z}_{3} = - 9 , {z}_{4} = - 1$
9) ${z}_{1} = - 9 , {z}_{2} = 1 , {z}_{3} = 0 , {z}_{4} = 2$
10) ${z}_{1} = - 9 , {z}_{2} = 1 , {z}_{3} = 9 , {z}_{4} = 1$
11) ${z}_{1} = - 6 , {z}_{2} = 0 , {z}_{3} = - 3 , {z}_{4} = - 3$
12) ${z}_{1} = - 6 , {z}_{2} = 0 , {z}_{3} = - 3 , {z}_{4} = 3$
13) ${z}_{1} = - 3 , {z}_{2} = - 3 , {z}_{3} = - 6 , {z}_{4} = 0$
14) ${z}_{1} = - 3 , {z}_{2} = - 3 , {z}_{3} = - 3 , {z}_{4} = 3$
15) ${z}_{1} = - 3 , {z}_{2} = - 3 , {z}_{3} = 0 , {z}_{4} = - 6$
16) ${z}_{1} = - 3 , {z}_{2} = - 3 , {z}_{3} = 3 , {z}_{4} = - 3$
17) ${z}_{1} = - 3 , {z}_{2} = 3 , {z}_{3} = - 6 , {z}_{4} = 0$
18) ${z}_{1} = - 3 , {z}_{2} = 3 , {z}_{3} = - 3 , {z}_{4} = - 3$
19) ${z}_{1} = - 3 , {z}_{2} = 3 , {z}_{3} = 0 , {z}_{4} = 6$
20) ${z}_{1} = - 3 , {z}_{2} = 3 , {z}_{3} = 3 , {z}_{4} = 3$
21) ${z}_{1} = - 2 , {z}_{2} = 0 , {z}_{3} = - 1 , {z}_{4} = - 9$
22) ${z}_{1} = - 2 , {z}_{2} = 0 , {z}_{3} = - 1 , {z}_{4} = 9$
23) ${z}_{1} = - 1 , {z}_{2} = - 9 , {z}_{3} = - 2 , {z}_{4} = 0$
24) ${z}_{1} = - 1 , {z}_{2} = - 9 , {z}_{3} = - 1 , {z}_{4} = 9$
25) ${z}_{1} = - 1 , {z}_{2} = - 9 , {z}_{3} = 0 , {z}_{4} = - 18$
26) ${z}_{1} = - 1 , {z}_{2} = - 9 , {z}_{3} = 1 , {z}_{4} = - 9$
27) ${z}_{1} = - 1 , {z}_{2} = 9 , {z}_{3} = - 2 , {z}_{4} = 0$
28) ${z}_{1} = - 1 , {z}_{2} = 9 , {z}_{3} = - 1 , {z}_{4} = - 9$
29) ${z}_{1} = - 1 , {z}_{2} = 9 , {z}_{3} = 0 , {z}_{4} = 18$
30) ${z}_{1} = - 1 , {z}_{2} = 9 , {z}_{3} = 1 , {z}_{4} = 9$
31) ${z}_{1} = 0 , {z}_{2} = - 18 , {z}_{3} = - 1 , {z}_{4} = - 9$
32) ${z}_{1} = 0 , {z}_{2} = - 18 , {z}_{3} = 1 , {z}_{4} = - 9$
33) ${z}_{1} = 0 , {z}_{2} = - 6 , {z}_{3} = - 3 , {z}_{4} = - 3$
34) ${z}_{1} = 0 , {z}_{2} = - 6 , {z}_{3} = 3 , {z}_{4} = - 3$
35) ${z}_{1} = 0 , {z}_{2} = - 2 , {z}_{3} = - 9 , {z}_{4} = - 1$
36) ${z}_{1} = 0 , {z}_{2} = - 2 , {z}_{3} = 9 , {z}_{4} = - 1$
37) ${z}_{1} = 0 , {z}_{2} = 2 , {z}_{3} = - 9 , {z}_{4} = 1$
38) ${z}_{1} = 0 , {z}_{2} = 2 , {z}_{3} = 9 , {z}_{4} = 1$
39) ${z}_{1} = 0 , {z}_{2} = 6 , {z}_{3} = - 3 , {z}_{4} = 3$
40) ${z}_{1} = 0 , {z}_{2} = 6 , {z}_{3} = 3 , {z}_{4} = 3$
41) ${z}_{1} = 0 , {z}_{2} = 18 , {z}_{3} = - 1 , {z}_{4} = 9$
42) ${z}_{1} = 0 , {z}_{2} = 18 , {z}_{3} = 1 , {z}_{4} = 9$
43) ${z}_{1} = 1 , {z}_{2} = - 9 , {z}_{3} = - 1 , {z}_{4} = - 9$
44) ${z}_{1} = 1 , {z}_{2} = - 9 , {z}_{3} = 0 , {z}_{4} = - 18$
45) ${z}_{1} = 1 , {z}_{2} = - 9 , {z}_{3} = 1 , {z}_{4} = 9$
46) ${z}_{1} = 1 , {z}_{2} = - 9 , {z}_{3} = 2 , {z}_{4} = 0$
47) ${z}_{1} = 1 , {z}_{2} = 9 , {z}_{3} = - 1 , {z}_{4} = 9$
48) ${z}_{1} = 1 , {z}_{2} = 9 , {z}_{3} = 0 , {z}_{4} = 18$
49) ${z}_{1} = 1 , {z}_{2} = 9 , {z}_{3} = 1 , {z}_{4} = - 9$
50) ${z}_{1} = 1 , {z}_{2} = 9 , {z}_{3} = 2 , {z}_{4} = 0$
51) ${z}_{1} = 2 , {z}_{2} = 0 , {z}_{3} = 1 , {z}_{4} = - 9$
52) ${z}_{1} = 2 , {z}_{2} = 0 , {z}_{3} = 1 , {z}_{4} = 9$
53) ${z}_{1} = 3 , {z}_{2} = - 3 , {z}_{3} = - 3 , {z}_{4} = - 3$
54) ${z}_{1} = 3 , {z}_{2} = - 3 , {z}_{3} = 0 , {z}_{4} = - 6$
55) ${z}_{1} = 3 , {z}_{2} = - 3 , {z}_{3} = 3 , {z}_{4} = 3$
56) ${z}_{1} = 3 , {z}_{2} = - 3 , {z}_{3} = 6 , {z}_{4} = 0$
57) ${z}_{1} = 3 , {z}_{2} = 3 , {z}_{3} = - 3 , {z}_{4} = 3$
58) ${z}_{1} = 3 , {z}_{2} = 3 , {z}_{3} = 0 , {z}_{4} = 6$
59) ${z}_{1} = 3 , {z}_{2} = 3 , {z}_{3} = 3 , {z}_{4} = - 3$
60) ${z}_{1} = 3 , {z}_{2} = 3 , {z}_{3} = 6 , {z}_{4} = 0$
61) ${z}_{1} = 6 , {z}_{2} = 0 , {z}_{3} = 3 , {z}_{4} = - 3$
62) ${z}_{1} = 6 , {z}_{2} = 0 , {z}_{3} = 3 , {z}_{4} = 3$
63) ${z}_{1} = 9 , {z}_{2} = - 1 , {z}_{3} = - 9 , {z}_{4} = - 1$
64) ${z}_{1} = 9 , {z}_{2} = - 1 , {z}_{3} = 0 , {z}_{4} = - 2$
65) ${z}_{1} = 9 , {z}_{2} = - 1 , {z}_{3} = 9 , {z}_{4} = 1$
66) ${z}_{1} = 9 , {z}_{2} = - 1 , {z}_{3} = 18 , {z}_{4} = 0$
67) ${z}_{1} = 9 , {z}_{2} = 1 , {z}_{3} = - 9 , {z}_{4} = 1$
68) ${z}_{1} = 9 , {z}_{2} = 1 , {z}_{3} = 0 , {z}_{4} = 2$
69) ${z}_{1} = 9 , {z}_{2} = 1 , {z}_{3} = 9 , {z}_{4} = - 1$
70) ${z}_{1} = 9 , {z}_{2} = 1 , {z}_{3} = 18 , {z}_{4} = 0$
71) ${z}_{1} = 18 , {z}_{2} = 0 , {z}_{3} = 9 , {z}_{4} = - 1$
72) ${z}_{1} = 18 , {z}_{2} = 0 , {z}_{3} = 9 , {z}_{4} = 1$

$- - - - - - - - - - - - - - - - - - -$

Step one:

$A = \left(0 , 0\right) , B = \left(b , 0\right) , C = \left(\frac{b}{2} , h\right)$

$| A C | = | B C |$

$9 = \frac{1}{2} \cdot b \cdot h R i g h t a r r o w h = \frac{18}{b}$

$C = \left(\frac{b}{2} , \frac{18 \epsilon}{b}\right)$

epsilon = ±1

Rotating by $\theta$,

$B = \left(\begin{matrix}b \cos \theta \\ b \sin \theta\end{matrix}\right)$

$C = \left(\begin{matrix}\frac{b}{2} \cos \theta - \frac{18 \epsilon}{b} \sin \theta \\ \frac{b}{2} \sin \theta + \frac{18 \epsilon}{b} \cos \theta\end{matrix}\right)$

$\left(b , \theta\right) \mapsto \left({z}_{1} , {z}_{2} , {z}_{3} , {z}_{4}\right) \in m a t h \boldsymbol{{Z}^{4}}$

$- - - - - - - - - - - - - - - - - - -$

Step two:

$A = \left(0 , 0\right) , B = \left(b , 0\right) , C = \left(b \cos t , b \sin t\right)$

$| A B | = | A C |$

$9 = \frac{1}{2} \cdot b \cdot \left(b \sin t\right) R i g h t a r r o w \frac{18}{\sin} t = {b}^{2} R i g h t a r r o w b = \frac{3 \sqrt{2}}{\sqrt{\sin}} t$

$C = 3 \sqrt{2} \left(\cos \frac{t}{\sqrt{\sin}} t , \epsilon \sqrt{\sin} t\right)$

epsilon = ± 1

I meant ${y}_{C}$ is a solution, then $- {y}_{C}$ is also a solution.

Rotating by $\theta$,

$B = 3 \sqrt{2} \left(\begin{matrix}\frac{1}{\sqrt{\sin}} t \cos \theta \\ \frac{1}{\sqrt{\sin}} t \sin \theta\end{matrix}\right)$

$C = 3 \sqrt{2} \left(\begin{matrix}\cos \frac{t}{\sqrt{\sin}} t \cos \theta - \epsilon \sqrt{\sin} t \sin \theta \\ \cos \frac{t}{\sqrt{\sin}} t \sin \theta + \epsilon \sqrt{\sin} t \cos \theta\end{matrix}\right)$

Again, $\left(t , \theta\right) \mapsto \left({z}_{5} , {z}_{6} , {z}_{7} , {z}_{8}\right) \in m a t h \boldsymbol{{Z}^{4}}$

$- - - - - - - - - - - - - - - - - -$

Step three:

Consider the ball centered at $B$ with radius $b$.

$A = \left(0 , 0\right) , B = \left(b , 0\right) , C = \left(b + b \cos t , b \sin t\right)$

$| B A | = | B C |$

$9 = \frac{1}{2} \cdot b \cdot \left(b \sin t\right) R i g h t a r r o w b = \frac{3 \sqrt{2}}{\sqrt{\sin}} t$

$C = 3 \sqrt{2} \left(\frac{1 + \cos t}{\sqrt{\sin}} t , \epsilon \sqrt{\sin} t\right)$

epsilon = ± 1

Rotating by $\theta$,

$B = 3 \sqrt{2} \left(\begin{matrix}\frac{1}{\sqrt{\sin}} t \cos \theta \\ \frac{1}{\sqrt{\sin}} t \sin \theta\end{matrix}\right)$

$C = 3 \sqrt{2} \left(\begin{matrix}\frac{1 + \cos t}{\sqrt{\sin}} t \cos \theta - \epsilon \sqrt{\sin} t \sin \theta \\ \frac{1 + \cos t}{\sqrt{\sin}} t \sin \theta + \epsilon \sqrt{\sin} t \cos \theta\end{matrix}\right)$

$\left(t , \theta\right) \mapsto \left({z}_{9} , {z}_{10} , {z}_{11} , {z}_{12}\right) \in m a t h \boldsymbol{{Z}^{4}}$ 