383] 
PROBLEMS AND SOLUTIONS. 
577 
Solution by the Proposer. 
Let 12, 45, meet in 0, and through 0 draw at pleasure a line meeting 14 in P, 
and 25 in Q; let P2, Q4 meet in 3, and P5, Q1 in 6; then the line 36 will pass 
through 0, and this being so, the hexagon 123456 satisfies the required conditions. 
We have to show that 36 passes through 0. Let Q4 meet 012 in A, and P2 
meet 045 in B; then the points 6, 3, 0, are the intersections of corresponding sides 
of the triangles A1Q, BoP; and in order that these points may lie in a line, the 
lines joining the corresponding vertices must meet in a point, that is, we have to show 
that the lines 15, AB, PQ meet in a point. The property is in fact as follows; viz., 
given the points 2, 4; and also the points Q, 0, P lying in a line; then constructing 
the points 1, 5, A, B, which are the respective intersections of P4, 02; Q2, 04; 
Q4, 02; P2, 04; the lines 15, AB, PQ will meet in a point. Take x = 0, y — 0, 
z = 0 for the respective equations of P‘2, Q4, PQ; then 0 is an arbitrary point in the 
line PQ, say that for the point 0 we have z = 0, ax + by = 0; also 02, 04 are 
arbitrary lines through 0 : say that their equations are ax + by + Xz = 0; ax+by + yz = 0; 
then we have for the points A and B, respectively, ax + by + yz = 0, y = 0; ax + by + yz = 0, 
x = 0; hence the equation of AB is yax + Xby+ \fiz = 0. The equation of P4 is 
ax + yz — 0, and that of Q2 is by + Xz = 0; the point 1 is therefore given by 
ax-\- yz = 0, ax + by + Xz = 0; and 5 by by 4- Xz = 0, ax+by + yz = 0; hence the equation 
of 15 is yax + Xby + (/A — + A, 2 ) z = 0 ; and the equation of PQ being z = 0, it is 
clear that the three lines AB, 15, PQ intersect in the point given by the equations 
fiax + \by = 0, z — 0. 
Obs. 1. By inspection of the figure we see that 3PQ is a triangle whereof the 
sides 3P, 3 Q, PQ pass respectively through the fixed points 2, 4, 0; while the vertices 
P and Q lie in the fixed lines 14, 25; the locus of the vertex 3 is consequently a 
conic; and the like as regards the triangle QPQ. 
Obs. 2. The regular hexagon projects into a hexagon inscribed in a conic and 
circumscribed about another conic having double contact therewith; in the hexagon 
so obtained (as appears at once by the consideration of the regular hexagon) the 
c. y. 73