integer triangle造句

例句与造句

1. All integer triangles with a 60?angle have their angles in an arithmetic progression.
2. Consequently, it can be stated that an integer triangle is Heronian if and only if it can be drawn as a lattice triangle.
3. The easiest way of generating lists of Heronian triangles is to generate all integer triangles up to a maximum side length and test for an integral area.
4. If the angles of any triangle form an arithmetic progression then one of its angles must be 60?. i . e . the integer triangle is equilateral.
5. Conditions are known in terms of elliptic curves for an integer triangle to have an integer ratio " N " of the circumradius to the inradius.
6. It's difficult to find integer triangle in a sentence. 用integer triangle造句挺难的
7. Thus the squared distance between the incenter and the circumcenter of an integer triangle, given by Euler's theorem as " R 2 & minus; 2Rr ", is rational.
8. Since all the terms under the radical on the right side of the formula are integers it follows that all integer triangles must have an integer value of " 16T 2 " and " T 2 " will be rational.
9. A "'rational triangle "'can be defined as one having all sides with rational length; any such rational triangle can be integrally rescaled ( can have all sides multiplied by the same integer, namely a common multiple of their denominators ) to obtain an integer triangle, so there is no substantive difference between integer triangles and rational triangles in this sense.
10. A "'rational triangle "'can be defined as one having all sides with rational length; any such rational triangle can be integrally rescaled ( can have all sides multiplied by the same integer, namely a common multiple of their denominators ) to obtain an integer triangle, so there is no substantive difference between integer triangles and rational triangles in this sense.
11. The square of each internal angle bisector of an integer triangle is rational, because the general triangle formula for the internal angle bisector of angle " A " is \ tfrac { 2 \ sqrt { bcs ( s-a ) } } { b + c } where " s " is the semiperimeter ( and likewise for the other angles'bisectors ).