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1. An acute-angled triangle lies in the plane such that the coordinates of its vertices
are all different integers and no sides are parallel to the coordinate axes. If the
triangle has area 348 and one side of length 29, what is the product of the lengths of the other two sides?

Any hints? I can't seem to get an algebraic approach to work as too many variables and results on lattice-point triangles e.g. Pick's Theorem seem promising but can't easily be related to side lengths.
2. (Original post by Prasiortle)
An acute-angled triangle lies in the plane such that the coordinates of its vertices
are all different integers and no sides are parallel to the coordinate axes. If the
triangle has area 348 and one side of length 29, what is the product of the lengths of the other two sides?

Any hints? I can't seem to get an algebraic approach to work as too many variables and results on lattice-point triangles e.g. Pick's Theorem seem promising but can't easily be related to side lengths.
Place one vertex at the origin and another 29 units from the origin. What could be the coordinates of this point? Now there is only one more point to find.
3. (Original post by BuryMathsTutor)
Place one vertex at the origin and another 29 units from the origin. What could be the coordinates of this point? Now there is only one more point to find.
The origin can't be a vertex as "the coordinates of its vertices are all different integers", whereas 0 and 0 (the x- and y-coordinates of the origin) are the same. In any case I tried placing one point at (-1, 0) and then another point at (19, 21), using the 20-21-29 Pythagorean triple, but I still don't see how to find the third point just from that.
4. (Original post by Prasiortle)
The origin can't be a vertex as "the coordinates of its vertices are all different integers", whereas 0 and 0 (the x- and y-coordinates of the origin) are the same. In any case I tried placing one point at (-1, 0) and then another point at (19, 21), using the 20-21-29 Pythagorean triple, but I still don't see how to find the third point just from that.
That requirement doesn't change the solution though.

Do you know how to find the area of a triangle using the cross product?
5. (Original post by BuryMathsTutor)
That requirement doesn't change the solution though.

Do you know how to find the area of a triangle using the cross product?
Yes, 1/2 * the magnitude of the cross product of the two vectors. But how does that help?
6. (Original post by Prasiortle)
Yes, 1/2 * the magnitude of the cross product of the two vectors. But how does that help?
Can you use it to find an equation to solve?
7. in the statement "the coordinates of its vertices
are all different integers and no sides are parallel to the coordinate axes" the highlighted part is automatically implied by the first part... you cannot have 3 pairs of coordinates where the 6 digits are different and have a side parallel to the x or the y axis
8. (Original post by BuryMathsTutor)
Can you use it to find an equation to solve?
If I let the third point have coordinates (a, b) and equate the area to 348, I can get an equation which could already have been reached much more easily using the Shoelace Formula. But then there's two variables and the only way to eliminate one is to attempt to incorporate the fact that the side lengths must all be integers...
9. (Original post by Prasiortle)
If I let the third point have coordinates (a, b) and equate the area to 348, I can get an equation which could already have been reached much more easily using the Shoelace Formula. But then there's two variables and the only way to eliminate one is to attempt to incorporate the fact that the side lengths must all be integers...
10. (Original post by BuryMathsTutor)
If we place points at (0,0), (21,20), and (a,b), we get 20a-21b=696.
11. (Original post by Prasiortle)
If we place points at (0,0), (21,20), and (a,b), we get 20a-21b=696.
Your equation is not quite correct. It does produce a solution (-3,-36) giving a triangle with the correct area but this is not the point that you require as this does not give an acute angled triangle.
12. (Original post by BuryMathsTutor)
Your equation is not quite correct. It does produce a solution (-3,-36) giving a triangle with the correct area but this is not the point that you require as this does not give an acute angled triangle.
Then I don't know how to get any other equation...
13. (Original post by Prasiortle)
Then I don't know how to get any other equation...
As you said, earlier, you need the magnitude..

21b-20a=696 gives the correct point.
14. (Original post by BuryMathsTutor)
As you said, earlier, you need the magnitude..

21b-20a=696 gives the correct point.
If the point is hence (3,36) then we get sides of 29, sqrt(1305), and sqrt(580), giving a product of 870, which is listed as the correct answer. Thanks for your assistance!
15. (Original post by Prasiortle)
If the point is hence (3,36) then we get sides of 29, sqrt(1305), and sqrt(580), giving a product of 870, which is listed as the correct answer. Thanks for your assistance!
You're welcome.

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