Solid O-ABCD in a figure is a right quadrangular pyramid.
The bottom is a square and all the length of OA, OB, OC and OD is equal.
Moreover, E and F divide OB and OD into the ratio of 3 : 1, respectively.
The intersection of the plane which passes along three point A, E, F and the side OC is set to G.
Find OG : GC.
Moreover, in two portions of the right quadrangular pyramid divided by this plane, find the volume ratio of the volume of a portion containing O and the volume of the right quadrangular pyramid.
The bottom is a square and all the length of OA, OB, OC and OD is equal.
Moreover, E and F divide OB and OD into the ratio of 3 : 1, respectively.
The intersection of the plane which passes along three point A, E, F and the side OC is set to G.
Find OG : GC.
Moreover, in two portions of the right quadrangular pyramid divided by this plane, find the volume ratio of the volume of a portion containing O and the volume of the right quadrangular pyramid.
Answer
Fig. 2 is a cross section which cut by the plane which passes along OBD.
Since bottom ABCD is a square, the line OH which connected the intersection H of AC and BD and the vertex O becomes vertical to both AC and BD.
The intersection of OH and AG is set to I in Fig. 1.
I is a point on EF in Fig. 2, and OF : FD = 3 : 1.
Thus, OI : IH = 3 : 1.
In Fig. 1, in order to calculate OG : GC, △GCJ is made by drawing an auxiliary line.
As for △ACJ, since AH : CH = 1 : 1, IH : JC = 1 : 2.
Considering that OI : IH = 3 : 1, it turns out OI : JC = 3 : 2.
Thus, △GOI and △GCJ are homothetic and a homothetic ratio is 3 : 2.
Therefore, OG : GC = 3 : 2.
The volume of the triangular pyramid E-OAG and the triangular pyramid B-AOC is measured.
The bottom of two triangular pyramids is set to △OAG and △OAC, respectively.
Since OG : GC = 3 : 2, the ratio of the area of △OAG to △OAC is 3 : (3+2) = 3 : 5.
The ratio of the height of two triangular pyramids turns into a ratio of EO to BO and is set to 3 : (3+1) = 3 : 4.
The triangular pyramid E-OAG will be the 3/5 × 3/4 = 9/20 times of the volume of the triangular pyramid B-AOC.
Completely think in the same way and the triangular pyramid F-OAG will also be 9/20 times of the volume of the triangular pyramid D-AOC.
Therefore, the volume ratio to be found is 9 : 20.
3 : 2
9 : 20
9 : 20
Solution
Fig. 1 is a cross section which cut right quadrangular pyramid by the plane which passes along OAC.
Fig. 2 is a cross section which cut by the plane which passes along OBD.
Since bottom ABCD is a square, the line OH which connected the intersection H of AC and BD and the vertex O becomes vertical to both AC and BD.
The intersection of OH and AG is set to I in Fig. 1.
I is a point on EF in Fig. 2, and OF : FD = 3 : 1.
Thus, OI : IH = 3 : 1.
In Fig. 1, in order to calculate OG : GC, △GCJ is made by drawing an auxiliary line.
As for △ACJ, since AH : CH = 1 : 1, IH : JC = 1 : 2.
Considering that OI : IH = 3 : 1, it turns out OI : JC = 3 : 2.
Thus, △GOI and △GCJ are homothetic and a homothetic ratio is 3 : 2.
Therefore, OG : GC = 3 : 2.
In order to compare the volume of the solid containing O with the volume of the right quadrangular pyramid, right quadrangular pyramid are divided into two at the plane OAC.
The volume of the triangular pyramid E-OAG and the triangular pyramid B-AOC is measured.
The bottom of two triangular pyramids is set to △OAG and △OAC, respectively.
Since OG : GC = 3 : 2, the ratio of the area of △OAG to △OAC is 3 : (3+2) = 3 : 5.
The ratio of the height of two triangular pyramids turns into a ratio of EO to BO and is set to 3 : (3+1) = 3 : 4.
The triangular pyramid E-OAG will be the 3/5 × 3/4 = 9/20 times of the volume of the triangular pyramid B-AOC.
Completely think in the same way and the triangular pyramid F-OAG will also be 9/20 times of the volume of the triangular pyramid D-AOC.
Therefore, the volume ratio to be found is 9 : 20.
