Problem:
The rectangle ABCD at the right has dimensions AB=123​ and BC=133​. Diagonals AC and BD intersect at P. If triangle ABP is cut out and removed, edges AP and BP are joined, and the figure is then creased along segments CP and DP, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
Solution:
Let m and n denote CD and BC, respectively. Three of the faces are lettered PBC, PCD, and PDB (see diagram). Let N be the point where the altitude from P meets BCD. We first show that N is the circumcenter of △BCD. To see this, note that △PNB,△PNC and △PND are congruent by the hypotenuse-leg criterion. It follows that BN,CN and DN all have the same length r; this r is the radius (and hence N is the center) of the circle that circumscribes △BCD. We will find the value of r and use it to find PN. From B draw a diameter of the circumcircle. Let the other end of this diameter be F and let E be the point where the diameter meets CD. Then ∠BFD≅∠ECB since both angles subtend the same arc on the circumcircle. Hence the two right triangles △BFD and △BCE are similar, implying BF/BD=BC/BE. Since BF=2r, the last equation gives
r=2(BE)n2​=4n2−m2​n2​
Now using PB=21​m2+n2​ we can find PN, the altitude of the pyramid, by the Pythagorean theorem:
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