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The final digit of a triangular number is 0, 1, 3, 5, 6, or 8, and thus such numbers never end in 2, 4, 7, or 9. A final 3 must be preceded by a 0 or 5; a final 8 must be preceded by a 2 or 7. In base 10, the digital root of a nonzero triangular number is always 1, 3, 6, or 9. Hence, every triangular number is either divisible by three or has a ...
If D > 1, no such triangle exists because the side b does not reach line BC. For the same reason a solution does not exist if the angle β ≥ 90° and b ≤ c. If D = 1, a unique solution exists: γ = 90°, i.e., the triangle is right-angled. If D < 1 two alternatives are possible. If b ≥ c, then β ≥ γ (the larger side corresponds to a ...
Whereas Roberts's theorem concerns the fewest possible triangles made by a given number of lines, the related Kobon triangle problem concerns the largest number possible. [3] The two problems differ already for n = 5 {\displaystyle n=5} , where Roberts's theorem guarantees that three triangles will exist, but the solution to the Kobon triangle ...
This equation can have 2, 1, or 0 positive solutions corresponding to the number of possible triangles given the data. It will have two positive solutions if b sin γ < c < b, only one positive solution if c = b sin γ, and no solution if c < b sin γ. These different cases are also explained by the side-side-angle congruence ambiguity.
After relating area to the number of triangles in this way, the proof concludes by using Euler's polyhedral formula to relate the number of triangles to the number of grid points in the polygon. [5] Tiling of the plane by copies of a triangle with three integer vertices and no other integer points, as used in the proof of Pick's theorem
The Kobon triangle problem is an unsolved problem in combinatorial geometry first stated by Kobon Fujimura (1903-1983). The problem asks for the largest number N(k) of nonoverlapping triangles whose sides lie on an arrangement of k lines.
All pairs of congruent triangles are also similar, but not all pairs of similar triangles are congruent. Given two congruent triangles, all pairs of corresponding interior angles are equal in measure, and all pairs of corresponding sides have the same length. This is a total of six equalities, but three are often sufficient to prove congruence ...
The same geometric strategy also works for triangles, as in the figure on the right: [4] if the large triangle has area 1, then the largest black triangle has area 1 / 4 , and so on. The figure as a whole has a self-similarity between the large triangle and its upper sub-triangle.