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?

2 Answers

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=(0,0) is a vertex of the triangle. We have a triangle ABO that has to be isosceles, so one of the following has to hold:

  • hat(A)=hat(O) and AB=BO
  • hat(B)=hat(O) and AB=AO
  • hat(A)=hat(B) and AO=BO

Condition [a] says that x_A, x_B, y_A, y_B in ZZ.

Condition [b] can be translated into a condition on the cross product between vec(OA) and vec(OB): S=1/2 ||vec(OA) times vec(OB)||=1/2 ||vec(OA)|| \ ||vec(OB)|| \ |sin hat(O) |=9 (where || cdot || is the magnitude of the vector and | cdot | is the absolute value). So sin hat(O)=pm 18/(AO \ BO) and if AO\ BO ge 18 (otherwise we get unacceptable cases) we conclude that hat(O)=arcsin( pm 18/(AO \ BO)).

So we translated the problem into mathematical language and we got limitation [1]: AO\ BO \ge 18.


FIRST MANIPULATIONS:

By the Pythagorean Theorem, the following identities hold:

  • AO^2=x_A^2+y_A^2
  • BO^2=x_B^2+y_B^2
  • AB^2=(x_A-x_B)^2+(y_A-y_B)^2

Note: by condition [a] we have that AO^2,BO^2,AB^2 in ZZ. This doesn't necessarily imply that AO,BO,AB in ZZ!

The last identity can be rewritten as follows (also known as Law of Cosines):
AB^2=AO^2+BO^2-2(x_Ax_B+y_A y_B)=AO^2+BO^2-2(vec(OA) cdot vec(OB))=AO^2+BO^2-2\ AO\ BO cos hat(O)

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


FIRST CASE:

We have three cases to analyze. Let's start with the first case: hat(A)=hat(O) and AB=BO.

The identity [2] becomes AO^2 = 2 sqrt{AO^2 \ BO^2-18^2}, so we can square the two (positive) sides and get
AO^4=4\ AO^2 \ BO^2-4 cdot 18^2
4\ AO^2 \ BO^2-AO^4=36^2
AO^2(4BO^2-AO^2)=2^4 * 3^4

By condition [a] we know that AO^2=x_A^2+y_A^2 in ZZ and 4BO^2-AO^2=4x_B^2+4y_B^2-x_A^2-y_A^2 in ZZ. 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. AO^2 is positive, so 4BO^2 - AO^2 ge 0 i.e. AO^2 le 4BO^2.
The solutions to the equation obtained above are among the following candidates: AO^2=2^k cdot 3^h and 4BO^2-AO^2=2^(4-k) cdot 3^(4-h) for h,k=0,1,2,3,4 (25 cases).

By condition [a] BO^2=x_B^2+y_B^2 in ZZ, then 4BO^2=2^(4-k) cdot 3^(4-h)+AO^2=2^(4-k) cdot 3^(4-h)+2^k cdot 3^h has to be divisible by 4.

  • If k=0, then 4BO^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 (4BO^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 4BO^2=2 cdot 3^(4-h)+2^3 cdot 3^h. It follows that 2BO^2=3^(4-h)+2^2 cdot 3^h is an addition between an odd and an even number, so the sum 2BO^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 AO^2=2^2 cdot 3^h and 4BO^2=2^2 cdot 3^(4-h)+2^2 cdot 3^h i.e. BO^2=3^(4-h)+3^h.

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

  • AO\ BO ge 18 iff AO^2 \ BO^2 ge 18^2 and substituting we get 3^4 + 3^(2h) ge 3^4 and this is satisfied for every h=0,1,2,3,4.
  • AO^2 le 4BO^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=AO^2=2^2*3^0=4 and 4=0^2+2^2, so we have (x_A,y_A)=(0,pm 2) or (x_A,y_A)=(pm 2,0); moreover x_B^2+y_B^2=BO^2=3^4+3^0=82 and 82=1^2+9^2, so we have (x_B,y_B)=(pm 1, pm 9) or (x_B,y_B)=(pm 9,pm 1).
  • h=1, then x_A^2+y_A^2=AO^2=2^2*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=AO^2=2^2*3^2=36 and 36=0^2+6^2, so we have (x_A,y_A)=(0,pm 6) or (x_A,y_A)=(pm 6,0); moreover x_B^2+y_B^2=BO^2=3^2+3^2=18 and 18=3^2+3^2, so we have (x_B,y_B)=(pm 3, pm 3).
  • h=3, then x_A^2+y_A^2=AO^2=2^2*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=AO^2=2^2*3^4=324 and 324=0^2+18^2, so we have (x_A,y_A)=(0,pm 18) or (x_A,y_A)=(pm 18,0); moreover x_B^2+y_B^2=BO^2=3^0+3^4=82 and 82=1^2+9^2, so we have (x_B,y_B)=(pm 1, pm 9) or (x_B,y_B)=(pm 9, pm 1).

So, considering that in this first case AB = BO, we get the following solutions:

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

SECOND CASE:

hat(B)=hat(O) and AB=BO.

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 AO=BO.

The identity [2] becomes
2AO^2-AB^2 = 2 sqrt{AO^4-18^2}, so we can square both sides under the condition that 2AO^2 ge AB^2 (otherwise solutions are not acceptable since they're not real) and get
(2AO^2-AB^2)^2=4(AO^4- 18^2)
4AO^4-4 AO^2 AB^2 + AB^4=4AO^4-36^2
AB^2(4AO^2-AB^2)=36^2
(4AO^2-AB^2)AB^2=2^4 cdot 3^4

AB^2=2^k * 3^h and 4AO^2-AB^2=2^(4-k) * 3^(4-h), so 4AO^2=2^k * 3^h+2^(4-k) * 3^(4-h) for h,k=0,1,2,3,4. Note that 4AO^2 ge AB^2 holds because we already requested that 2AO^2 ge AB^2.
Since AO^2 in ZZ, by the same argument used in the first case for BO in place of AB, we conclude that 4AO^2 is divisible by 4 if and only if k=2. So AB^2=4 * 3^h, 4AO^2-AB^2=4 * 3^(4-h) and AO^2=3^h+3^(4-h).

We have to check two conditions:

  • AO^2 ge 18 iff 3^h+3^(4-h) ge 18 and this is satisfied for each h=0,1,2,3,4.
  • 2AO^2 ge AB^2 iff 2 cdot 3^h+2 cdot 3^(4-h) ge 4 * 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=AO^2=3^0+3^4=82 and 82=1^2+9^2, so we have (x_A,y_A)=(pm 1, pm 9) or (x_A,y_A)=(pm 9, pm 1); moreover (x_A-x_B)^2+(y_A-y_B)^2=AB^2=4 cdot 3^0=4 and 4=0+2^2, so we have (x_A-x_B,y_A-y_B)=(0,pm 2) or (x_A-x_B,y_A-y_B)=(pm 2,0) (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=AO^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=AO^2=3^2+3^2=18 and 18=3^3+3^2, so we have (x_A,y_A)=(pm 3,pm 3); moreover (x_A-x_B)^2+(y_A-y_B)^2=AB^2=4 cdot 3^2=36 and 36=0+6^2, so we have (x_A-x_B,y_A-y_B)=(0,pm 6) or (x_A-x_B,y_A-y_B)=(pm 6,0).

So, considering that in this third case AO = BO, we get the following solutions:

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

CONCLUSION

We proved that there are 24+12=36 different isosceles triangles ABO 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 [-18, 18] cap mathbb{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 = (0, 0), B = (b, 0), C = (b/2, h)

|AC| = |BC|

9 = 1/2 * b * h Rightarrow h = 18/b

C = (b/2, (18epsilon)/b)

epsilon = ±1

Rotating by theta,

B = ((b cos theta), (b sin theta))

C = ((b/2 cos theta - (18 epsilon)/b sin theta), (b/2 sin theta + (18 epsilon)/b cos theta))

(b, theta) mapsto (z_1, z_2, z_3, z_4) in mathbb{Z^4}

-------------------

Step two:

A = (0,0), B = (b, 0), C = (b cos t, b sin t)

|AB| = |AC|

9 = 1/2 * b * (b sin t) Rightarrow 18/sin t = b^2 Rightarrow b = (3 sqrt 2) / sqrt sin t

C = 3 sqrt 2 (cos t / sqrt sin t, epsilon sqrt sin t)

epsilon = ± 1

I meant y_C is a solution, then -y_C is also a solution.

Rotating by theta,

B = 3 sqrt 2((1/sqrt sin t cos theta), (1/sqrt sin t sin theta))

C = 3 sqrt 2 ((cos t / sqrt sin t cos theta - epsilon sqrt sin t sin theta), (cos t / sqrt sin t sin theta + epsilon sqrt sin t cos theta))

Again, (t, theta) mapsto (z_5, z_6, z_7, z_8) in mathbb{Z^4}

------------------

Step three:

Consider the ball centered at B with radius b.

A = (0, 0), B = (b, 0), C = (b + b cos t, b sin t)

|BA| = |BC|

9 = 1/2 cdot b cdot (b sin t) Rightarrow b = (3 sqrt 2)/sqrt sin t

C = 3 sqrt 2 ((1 + cos t)/sqrt sin t, epsilon sqrt sin t)

epsilon = ± 1

Rotating by theta,

B = 3 sqrt 2 ((1/sqrt sin t cos theta), (1/sqrt sin t sin theta))

C = 3 sqrt 2 (((1 + cos t)/sqrt sin t cos theta - epsilon sqrt sin t sin theta), ((1 + cos t)/sqrt sin t sin theta + epsilon sqrt sin t cos theta))

(t, theta) mapsto (z_9, z_10, z_11, z_12) in mathbb{Z^4}

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