TOPICS
Search

Search Results for ""


671 - 680 of 1567 for Pascal's TriangleSearch Results
257 is a Fermat prime, and the 257-gon is therefore a constructible polygon using compass and straightedge, as proved by Gauss. An illustration of the 257-gon is not included ...
The regular hexagon is the regular polygon with six sides, as illustrated above. The inradius r, circumradius R, sagitta s, and area A of a regular hexagon can be computed ...
Given a circle expressed in trilinear coordinates by a central circle is a circle such that l:m:n is a triangle center and k is a homogeneous function that is symmetric in ...
The point of concurrence of the four maltitudes of a cyclic quadrilateral. Let M_(AC) and M_(BD) be the midpoints of the diagonals of a cyclic quadrilateral ABCD, and let P ...
Given two bicentric points P=p:q:r and U=u:v:w, their bicentric sum is defined by p+u:q+v:r:w.
One name for the figure used by Euclid to prove the Pythagorean theorem. It is sometimes also known as the "windmill."
Extend the symmedians of a triangle DeltaA_1A_2A_3 to meet the circumcircle at P_1, P_2, P_3. Then the symmedian point K of DeltaA_1A_2A_3 is also the symmedian point of ...
Let P=p:q:r and U=u:v:w be distinct trilinear points, neither lying on a sideline of DeltaABC. Then the crossdifference of P and U is the point X defined by trilinears ...
In the above figure, let E be the intersection of AD and BC and specify that AB∥EF∥CD. Then 1/(AB)+1/(CD)=1/(EF). A beautiful related theorem due to H. Stengel can be stated ...
The Euler infinity point is the intersection of the Euler line and line at infinity. Since it lies on the line at infinity, it is a point at infinity. It has triangle center ...
1 ... 65|66|67|68|69|70|71 ... 157 Previous Next

...