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
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
#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
Condition [b] can be translated into a condition on the cross product between
So we translated the problem into mathematical language and we got limitation [1]:
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
The last identity can be rewritten as follows (also known as Law of Cosines):
Substituting
This can be rewritten as the identity [2]
FIRST CASE:
We have three cases to analyze. Let's start with the first case:
The identity [2] becomes
By condition [a] we know that
The solutions to the equation obtained above are among the following candidates:
By condition [a]
- 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
#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 (
#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
#A=(0,2)# and#B=(9,1)# #A=(0,2)# and#B=(-9,1)# #A=(0,-2)# and#B=(9,-1)# #A=(0,-2)# and#B=(-9,-1)# #A=(2,0)# and#B=(1,9)# #A=(2,0)# and#B=(1,-9)# #A=(-2,0)# and#B=(-1,9)# #A=(-2,0)# and#B=(-1,-9)# #A=(0,6)# and#B=(3,3)# #A=(0,6)# and#B=(-3,3)# #A=(0,-6)# and#B=(3,-3)# #A=(0,-6)# and#B=(-3,-3)# #A=(6,0)# and#B=(3,3)# #A=(6,0)# and#B=(3,-3)# #A=(-6,0)# and#B=(-3,3)# #A=(-6,0)# and#B=(-3,-3)# #A=(0,18)# and#B=(1,9)# #A=(0,18)# and#B=(-1,9)# #A=(0,-18)# and#B=(1,-9)# #A=(0,-18)# and#B=(-1,-9)# #A=(18,0)# and#B=(9,1)# #A=(18,0)# and#B=(9,-1)# #A=(-18,0)# and#B=(-9,1)# #A=(-18,0)# and#B=(-9,-1)#
SECOND CASE:
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
THIRD CASE:
The identity [2] becomes
Since
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
#A=(1,9)# and#B=(1,-9)# #A=(1,9)# and#B=(-1,9)# #A=(-1,9)# and#B=(-1,-9)# #A=(-1,-9)# and#B=(1,-9)# #A=(9,1)# and#B=(9,-1)# #A=(9,1)# and#B=(-9,1)# #A=(9,-1)# and#B=(-9,-1)# #A=(-9,-1)# and#B=(-9,1)# #A=(3,3)# and#B=(3,-3)# #A=(3,3)# and#B=(-3,3)# #A=(-3,3)# and#B=(-3,-3)# #A=(-3,-3)# and#B=(3,-3)#
CONCLUSION
We proved that there are
Maybe it's easier to check all possibilities of four integers in
Explanation:
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
18)
19)
20)
21)
22)
23)
24)
25)
26)
27)
28)
29)
30)
31)
32)
33)
34)
35)
36)
37)
38)
39)
40)
41)
42)
43)
44)
45)
46)
47)
48)
49)
50)
51)
52)
53)
54)
55)
56)
57)
58)
59)
60)
61)
62)
63)
64)
65)
66)
67)
68)
69)
70)
71)
72)
Step one:
Rotating by
Step two:
I meant
Rotating by
Again,
Step three:
Consider the ball centered at
Rotating by
