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

Answer:

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

Answer:

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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