248 WARREN J. MEAD 
a fourth rock as possible. An apparatus was constructed embody- 
ing the ideas shown in Fig. 2 and used in the solution of the problem. 
Silk threads were used for the edges and intersections of the planes, 
and to avoid confusion a different color was used for each component. 
It was found, however, that more accurate results could be obtained 
by the following application of the principles involved in the appara- 
tus. The three vertical planes are opened out flat, as shown in Fig. 4. 
The percentages are plotted as before, and intersections found at 
X’ and Y’ and projected down upon the base-line. The distances 
XY, YZ, and ZX correspond to the sides of the triangle, so the loca- 
tion of the points x and y is an easy matter. The problem admits 
of easy solution by this graphical method, but a very simple arithmetical 
solution is also possible which does away with the rather cumbersome 
details of a graphical solution and also admits of greater accuracy. 
In Fig. 5 the lines AB and A’B’ are drawn parallel to the base. 
In the left-hand side of the figure we have two similar triangles, 
ABC and AOx. HencelG 5 AB <110 740; 
CB=40—10=30, x/O=25—I0=15, 
AB is equal in length to the sides of the triangle which is divided into 
100 divisions, hence AB =1Ioo. 
Substituting the values in the above proportion, we have, 
BO LOO seth aA). 
From which, 4O =50, and we see that the point x’ falls on the 
side X’Y’ of the triangle at a distance of 50 units from X’. 
Considering the right-hand side of the figure, we have as before 
two similar triangles, C’B’A’ and y’O’A’, and we may write the fol- 
lowing proportion, 
CB BAL sy OR .OtAS, 
C’B’=60—I10=50, y’O'’=25—10=15, B’A’=100, 
substituting values in the above proportion, 
50) cLOONs 1G) OL ACR 
solving, 
O'A'=30. 
We find that the point y’’ falls on the side X’Z’ of the triangle at a 
distance of 30 units from X’. 
