Time : 60 minutes
Passing mark : 70%
Answer : End of the problem
Problem 1
Problem 2
Problem 3
Problem 6
Find X.
1/2009 + 1/392 = 1/X
Problem 2
When 180 pencils were distributed to all students of A class per four pencils per person, several remained.
When another 200 pencils are distributed to all students of B class per six pencils per person, several fall short.
Then, when five pencils per person were distributed in B class, several remained.
Moreover, all remaining pencils in both classes could be exactly distributed to all students of A and B class per one pencil per person and no pencil remained.
Find the number of the student of A class.
Problem 3
There is a hexagon ABCDEF as shown in a figure.
How many ways of method of the division which divides this hexagon into three parts by two diagonal lines is there in all?
Problem 4
How many ways of method of the division which divides this hexagon into three parts by two diagonal lines is there in all?
Problem 4
There is a straight road connecting A point and B point.
Taro walks from A point to B point and Jiro walks from B point to A point with a fixed speed respectively.
Both of them started at the same time and passed on the way.
25 minutes after passing, Taro arrived at B point and Jiro arrived at A point the 24 minutes after Taro’s arrival.
Find the least integer ratio of the speed of Taro and Jiro.
Problem 5
There is a disk as shown in a figure and three needles which continue rotating with a fixed speed clockwise respectively around the center O of the disk.
The time concerning a needle rotating one time is 5 minutes, 8 minutes, and 14 minutes sequentially from a long needle.
All of three needles overlapped at a certain time.
Find the time concerning all of three needles overlapping next.
The time concerning a needle rotating one time is 5 minutes, 8 minutes, and 14 minutes sequentially from a long needle.
All of three needles overlapped at a certain time.
Find the time concerning all of three needles overlapping next.
Problem 6
(1) Find the number of an integer more than or equal to 1 and less than or equal to 999 which are not a multiple of 9 and do not contain 9 in the number of each digit.
(2) For the integer applicable to (1), find the 999th number counting from the least one in all integers.
Problem 7
The following operation is repeated for a certain integer repeatedly.
<Operation>
It doubles.
However,
When the doubled number is 150 or more, 100 is subtracted from this doubled number.
When the doubled numbers is 101 or more and 149 or less, 50 is subtracted from this doubled number.
When the doubled number is 100 or less, it leaves in this doubled number.
For example, when this operation is repeated 4 times starting with 36, the integer acquired is 72, 94, 88, and 76 at order.
(1) When you repeat this operation 4 times, find the number of integers that the result becomes 60.
(2) When you repeat this operation 101 times, find the least number among the integers that the result becomes 60.
<Operation>
It doubles.
However,
When the doubled number is 150 or more, 100 is subtracted from this doubled number.
When the doubled numbers is 101 or more and 149 or less, 50 is subtracted from this doubled number.
When the doubled number is 100 or less, it leaves in this doubled number.
For example, when this operation is repeated 4 times starting with 36, the integer acquired is 72, 94, 88, and 76 at order.
(1) When you repeat this operation 4 times, find the number of integers that the result becomes 60.
(2) When you repeat this operation 101 times, find the least number among the integers that the result becomes 60.
Problem 8
There is a parallelogram ABCD as shown in a figure.
The points E and F are points of dividing the side BC into three equally.
The point G is a point of the middle of the side CD.
Find the area ratio of the area of a shadow area and the area of parallelogram ABCD.
Problem 9
The points E and F are points of dividing the side BC into three equally.
The point G is a point of the middle of the side CD.
Find the area ratio of the area of a shadow area and the area of parallelogram ABCD.
Problem 9
As shown in a figure, the circle A centering on the point A with a radius of 3 cm and the circle B centering on the point B with a radius of 2 cm have touched.
The points C and D are points on the circumference of the circle B.
The points E and F are points on the circumference of the circle A.
Three point A, E and C are on a straight line and three point A, F and D are also on a straight line.
Find the area ratio of a quadrangle ACBD and the triangle AEF.
Problem 10
Fig. 1 shows the figure which put two disks 1 cm in radius together at the point A.
Place this figure into the frame of the right triangle of Fig. 2 and when you move this figure so as not to protrude from the frame, find the area of the range where the point A can move.
The points C and D are points on the circumference of the circle B.
The points E and F are points on the circumference of the circle A.
Three point A, E and C are on a straight line and three point A, F and D are also on a straight line.
Find the area ratio of a quadrangle ACBD and the triangle AEF.
Problem 10
Place this figure into the frame of the right triangle of Fig. 2 and when you move this figure so as not to protrude from the frame, find the area of the range where the point A can move.
---------------------------------------------------------------------------------------------
Problem 11
The straight line L of the figure divides the shadow area into three parts with an equal area.
Problem 12
Problem 13
Find the length of AB.
Problem 12
As shown in Fig.1, there are three point A, B and P on the even ground.
There are ABCD of a 3m-high rectangular wall and PQ of a 9m-high pillar standing on the ground straight.
Fig.2 shows the picture looked at these from right above.
Wall ABCD is illuminated by the electric light in the position at the tip Q of the pillar.
Find the area of the shadow of the wall made into the ground.
Neither the size of an electric light nor the thickness of a wall shall be considered.
There are ABCD of a 3m-high rectangular wall and PQ of a 9m-high pillar standing on the ground straight.
Fig.2 shows the picture looked at these from right above.
Wall ABCD is illuminated by the electric light in the position at the tip Q of the pillar.
Find the area of the shadow of the wall made into the ground.
Neither the size of an electric light nor the thickness of a wall shall be considered.
Problem 13
As for quadrangular pyramid O-ABCD in a figure, the bottom is a square and all the length of OA, OB, OC and OD is equal. K, L, M, N, P, Q, R, and S are the middle points of each side.
This quadrangular pyramid is cut by three planes of the plane which passes along P, K, N, R and the plane which passes along P, L, M, R and the plane which passes along S, K, L, Q and divides into some solids.
Find the ratio of the volume of a solid including the vertex O and the volume of the quadrangular pyramid O-ABCD.
This quadrangular pyramid is cut by three planes of the plane which passes along P, K, N, R and the plane which passes along P, L, M, R and the plane which passes along S, K, L, Q and divides into some solids.
Find the ratio of the volume of a solid including the vertex O and the volume of the quadrangular pyramid O-ABCD.
Answer
Problem 1
Problem 2
Problem 3
Find X.
1/2009 + 1/392 = 1/X
Answer
328
Problem 2
When 180 pencils were distributed to all students of A class per four pencils per person, several remained.
When another 200 pencils are distributed to all students of B class per six pencils per person, several fall short.
Then, when five pencils per person were distributed in B class, several remained.
Moreover, all remaining pencils in both classes could be exactly distributed to all students of A and B class per one pencil per person and no pencil remained.
Find the number of the student of A class.
Answer
34 persons
Solution
According to the text of the problem, sum total of remaining pencils after distribution both in A and B classes is equal to the number of students of A class and B class.
That is, 180 + 200 = 380 pencils can be distributed 4 +1 = 5 pencils per person to all A class students and 5 + 1 = 6 pencils per person to all B class students and there is no remainder.
When this is denoted by a formula, it is 5 × A + 6 × B = 380 (A,B express number of students).
Moreover, 200 / 6 = 33.3 --- and 200 / 5 = 40.
Thereby, the numbers of B class are 34 or more persons and 39 persons or less.
The number which is applied to the upper formula is 35.
Therefore, the number of A group is (380 - 6 × 35) / 5 = 34 persons.
Problem 3
There is a hexagon ABCDEF as shown in a figure.
How many ways of method of the division which divides this hexagon into three parts by two diagonal lines is there in all?
How many ways of method of the division which divides this hexagon into three parts by two diagonal lines is there in all?
Answer
Problem 4
21 ways
Solution
There are two cases to divide a hexagon into three parts by two diagonal lines as follows.
① Draw two diagonal lines from the one vertex
② Draw two diagonal lines not crossing from different vertex
① Three diagonal lines can be drawn from one vertex.
① Draw two diagonal lines from the one vertex
② Draw two diagonal lines not crossing from different vertex
① Three diagonal lines can be drawn from one vertex.
The number of ways of drawing two diagonal lines from one vertex turns into the number of ways of choosing two lines among three lines.
It is 3 × 2 / (2 × 1) = 3 ways.
Since there are six vertex,there are 3 × 6 = 18 ways in all.
② There is three groups of the diagonal lines of AC and FD, AE and BD, BF and CE.
The sum total above is 18 + 3 = 21 ways.
Problem 4
There is a straight road connecting A point and B point.
Taro walks from A point to B point and Jiro walks from B point to A point with a fixed speed respectively.
Both of them started at the same time and passed on the way.
25 minutes after passing, Taro arrived at B point and Jiro arrived at A point the 24 minutes after Taro’s arrival.
Find the least integer ratio of the speed of Taro and Jiro.
Answer
7 : 5
Solution
The time taken from the start until passing is assumed □.
Since Jiro took 25 + 24 = 49 minutes to walk the same distance Taro walked in □ minutes,
the ratio of the time to walk the same distance is Taro : Jiro = □ : 49 -------A.
Next, Since Jiro took □ minutes to walk the same distance Taro walked in 25 minutes,
the ratio of the time to walk the same distance is Taro : Jiro = 25 : □ -------B.
Since the ratio of the time to walk the same distance becomes the same, the proportional expression of A is equal to B.
□ : 49 = 25 : □.
49 × 25 = □ × □
49 × 25 = 7 × 7 × 5 × 5 = 7 × 5 × 7 × 5
Thus, □ = 7 × 5 = 35.
□ : 49 = 25 : □.
49 × 25 = □ × □
49 × 25 = 7 × 7 × 5 × 5 = 7 × 5 × 7 × 5
Thus, □ = 7 × 5 = 35.
Therefore the ratio of time to walk the same distance is Taro : Jiro = 35 : 49 = 5 : 7.
Since the ratio of the speed turns into an inverse ratio of the time ratio,
Since the ratio of the speed turns into an inverse ratio of the time ratio,
the speed ratio is Taro : Jiro = 7 : 5.
Problem 5
There is a disk as shown in a figure and three needles which continue rotating with a fixed speed clockwise respectively around the center O of the disk.
Problem 6
The time concerning a needle rotating one time is 5 minutes, 8 minutes, and 14 minutes sequentially from a long needle.
All of three needles overlapped at a certain time.
Find the time concerning all of three needles overlapping next.
Answer
850/9 minutes
Solution
Three needles are set to A, B, and C from the longer one in order.
Time to find is the time concerning two cycles between overlapping of A and B and overlapping of A and C come to be same.
Find the angle which A, B, and C advances in one minute respectively.
A : 360 degree / 5 minutes = 72 degrees/minute
B : 360 degree / 8 minutes = 45 degrees/minute
C : 360 degree / 14 minutes = 180/7 degrees/minute
A : 360 degree / 5 minutes = 72 degrees/minute
B : 360 degree / 8 minutes = 45 degrees/minute
C : 360 degree / 14 minutes = 180/7 degrees/minute
The time cycle A and B overlaps is the time concerning A catching up with B and it is 360 /(72 - 45) = 40/3 minutes.
The time cycle A and C overlaps is the time concerning A catching up with C and it is 360 / (72 - 180/7) = 709 minutes.
The time concerning two cycle overlap should be found by calculating the least common multiple of 40/3 and 70/9.
The time to find is 850/9 minutes.
Problem 6
(1) Find the number of an integer more than or equal to 1 and less than or equal to 999 which are not a multiple of 9 and do not contain 9 in the number of each digit.
(2) For the integer applicable to (1), find the 999th number counting from the least one in all integers.
Answer
(1) 648 pieces
(2) 1477
(2) 1477
Solution
(1)
The number suitable for conditions is counted sequentially from 1.
In 1, 2, 3, 4, 5, 6, 7, 8, 9, there are eight other than 9.
In 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, there are eight other than 18 and 19.
In 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, there are eight other than 27 and 29.
In 30 - 39, there are eight other than 36 and 39.
In 40 - 49, 50 - 59, 60 - 69, 70 - 79, 80 - 89, there are also eight pieces respectively.
In 90 - 99, there is zero.
It turns out that there are in total 8 pieces × 9 = 72 pieces from 1 to 99.
As for 100 - 199, since it is necessary to investigate ones digit and tens digit only, there are also 72 pieces as well as in 1 - 99.
Since 200 - 899 is similarly completely considered, it will be 72 pieces × 9 = 648 pieces in 1 - 899.
In 1, 2, 3, 4, 5, 6, 7, 8, 9, there are eight other than 9.
In 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, there are eight other than 18 and 19.
In 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, there are eight other than 27 and 29.
In 30 - 39, there are eight other than 36 and 39.
In 40 - 49, 50 - 59, 60 - 69, 70 - 79, 80 - 89, there are also eight pieces respectively.
In 90 - 99, there is zero.
It turns out that there are in total 8 pieces × 9 = 72 pieces from 1 to 99.
As for 100 - 199, since it is necessary to investigate ones digit and tens digit only, there are also 72 pieces as well as in 1 - 99.
Since 200 - 899 is similarly completely considered, it will be 72 pieces × 9 = 648 pieces in 1 - 899.
There is no integer suitable for the condition in 900 - 999.
The number to be found is 648 pieces.
(2)
(2)
Select numbers suitable for the condition sequentially in the same way as (1), there are 72 pieces in 1000 - 1099.
There are 72 in 1100 - 1199, 1200 - 1299, 1300 - 1399.
It becomes 72 × 13 = 936 in 1-1399 in total.
It becomes 72 × 13 = 936 in 1-1399 in total.
The remainder is 999-936 = 63.
It becomes eight in 1410-1419 and in 1400-1409 ---------.
According to the calculation of 63 / 8 = 7 remainder 7, the seventh number from 1470 is 1477 (1470, 1471, 1472, 1473, 1474, 1475, 1477).
It becomes eight in 1410-1419 and in 1400-1409 ---------.
According to the calculation of 63 / 8 = 7 remainder 7, the seventh number from 1470 is 1477 (1470, 1471, 1472, 1473, 1474, 1475, 1477).
The 999th number to find is 1477.
Problem 7
The following operation is repeated for a certain integer repeatedly.
<Operation>
It doubles.
However,
When the doubled number is 150 or more, 100 is subtracted from this doubled number.
When the doubled numbers is 101 or more and 149 or less, 50 is subtracted from this doubled number.
When the doubled number is 100 or less, it leaves in this doubled number.
For example, when this operation is repeated 4 times starting with 36, the integer acquired is 72, 94, 88, and 76 at order.
<Operation>
It doubles.
However,
When the doubled number is 150 or more, 100 is subtracted from this doubled number.
When the doubled numbers is 101 or more and 149 or less, 50 is subtracted from this doubled number.
When the doubled number is 100 or less, it leaves in this doubled number.
For example, when this operation is repeated 4 times starting with 36, the integer acquired is 72, 94, 88, and 76 at order.
(1) When you repeat this operation 4 times, find the number of integers that the result becomes 60.
(2) When you repeat this operation 101 times, find the least number among the integers that the result becomes 60.
Answer
(1) Four (10, 35, 60, 85)
(2) 5
(2) 5
Solution
(1) It goes back in good order and thinks.
Since it is a number which doubles and becomes 160, 110, and 60, it is 80, 55, or 30 that becomes 60 after the 4th operation.
Since there is no number which doubles and becomes 55, 55 is not suitable as a number after the 3rd operation.
55 is excepted. It is 90 and 40 that becomes 80 or 30 after the 3rd operation.
It is 70 and 20 that becomes 90 or 40 after the 2nd operation.
Furthermore, since it is 85, 60, 35, and 10 that becomes 70 or 20 after the 1st operation.
Therefore the number of suitable integer is four.
(2) Following (1), when the number becomes 60 by five operations is picked up, they are four numbers, 80, 55, 30, and 5.
When the number becomes 60 by six operations is also picked up, it is 90, 65, 40, and 15.
The result of (1) shows 60 turning into 60 by four operations and being set to 60 by repetition of four operations.
Then, according to the calculation of 101 / 4 = 25 ... 1, the number suitable as an answer is in numbers which becomes 60 by one or five operation.
Since the numbers becomes 60 by 1 time or 5 times are 80, 55, 30, and 5, the least number in these is 5.
Problem 8
There is a parallelogram ABCD as shown in a figure.
The points E and F are points of dividing the side BC into three equally.
The point G is a point of the middle of the side CD.
Find the area ratio of the area of a shadow area and the area of parallelogram ABCD.
Problem 9
The points E and F are points of dividing the side BC into three equally.
The point G is a point of the middle of the side CD.
Find the area ratio of the area of a shadow area and the area of parallelogram ABCD.
Answer
5 : 18
Solution
An auxiliary line HG is drawn as shown in Fig.1.
The area of △AJG of a shadow area is GI × D'J' / 2.
When it is set that BC = 3 and D'C ' = 2, the area of a parallelogram is 3 × 2 = 6.
HI = BF × 1/2 = 2 × 1/2 = 1.
GI =3 - 1 = 2.
△GIJ and △EFJ are homothetic and a homothetic ratio is 2 : 1.
When GI is set to the base of △GIJ, the height of △GIJ is G'J' = 1 × 2/(1+2) = 2/3.
Thus, △AJG = GI × D'J' / 2 = 2 × (1+ 2/3) / 2 = 5/3.
Therefore, △AJG : Parallelogram = 5/3 : 6 = 5 : 18.
Problem 9
As shown in a figure, the circle A centering on the point A with a radius of 3 cm and the circle B centering on the point B with a radius of 2 cm have touched.
The points C and D are points on the circumference of the circle B.
The points E and F are points on the circumference of the circle A.
Three point A, E and C are on a straight line and three point A, F and D are also on a straight line.
Find the area ratio of a quadrangle ACBD and the triangle AEF.
The points C and D are points on the circumference of the circle B.
The points E and F are points on the circumference of the circle A.
Three point A, E and C are on a straight line and three point A, F and D are also on a straight line.
Find the area ratio of a quadrangle ACBD and the triangle AEF.
Answer
Problem 10
25 : 9
Solution
Since the length of AB is equal to the sum length of each radius of the circle A and the circle B, it is 3 + 2 = 5cm.
△ABD and △AFG are homothetic and a homothetic ratio is AB : AF = 5 : 3.
An area ratio is 5 × 5 : 3 × 3 = 25 : 9.
The area of △AEF is twice of △AFG and the area of a quadrangle ACBD is also the twice of △ABD.
The area ratio of a quadrangle ACBD and △AEF is also 25 : 9.
Problem 10
Fig. 1 shows the figure which put two disks 1 cm in radius together at the point A.
Place this figure into the frame of the right triangle of Fig. 2 and when you move this figure so as not to protrude from the frame, find the area of the range where the point A can move.
Answer
4.43 cm2
Solution
Each center of two circles of Fig. 1 is set to P and Q.
When the circles P and Q come to the angle of the vertex B, it looks as it is shown in Fig. 3.
The circle Q moves to Q - Q' so that the arc with a radius of PQ as shown in a figure may be drawn, and the point A also moves to A' on the circumference of the circle P.
These two circles move in a same motion also on the corner of the vertex A and C.
Thus, the range where the point A can move becomes a shadow area of Fig. 4.
The area to determine becomes the shadow area which the areas of three sectors of the circle P, Q, and R of Fig. 4 are subtracted from the area of △PQR.
Each radius of three sectors is same 1 cm and the sum of central angles of three sectors is same as the sum of angles of △PQR as 180°.
The area is 1 × 1 × 3.14 × 180/360 = 1.57 cm2.
Since △PQR and △ABC are homothetic, the length of the side PQ is to be found first.
In Fig. 4, since PQ = DE, the length of DE is to be found. The length of DE is BC - BD - CE.
As shown in Fig. 5, the intersection of the line lengthen PD and the side AB is set to S.
As shown in Fig. 5, the intersection of the line lengthen PD and the side AB is set to S.
The point which the circle P touches the side AB is set to T.
Since △PTS, △SBD and △ABC have each of three equal angles, they are homothetic.
As for △PTS, PT : PS = BC : AB = 8 : 10 = 4 : 5.
Thus, PS = PT × 5/4 = 1 × 5/4 = 5/4 cm.
SD = 5/4 + 1 = 9/4 cm.
Since SD : BD = AC : BC = 6 : 8 = 3 : 4, it turns out to be BD = SD × 4/3 = 9/4 × 4/3 = 3cm.
Thus, PQ = DE = BC - 3cm - 1cm =8 - 3 - 1 = 4 cm.
Moreover, since RQ = PQ × 3/4 = 4 × 3/4 = 3 cm, the area of △PQR = 4 × 3 / 2 = 6 cm2.
Therefore, the area to be found is 6 - 1.57 = 4.43 cm2.
Problem 11
The straight line L of the figure divides the shadow area into three parts with an equal area.
Find the length of AB.
Answer
Problem 12
13/3 cm
Solution
The total area of this figure is 18 × 8 - 14 × 6 = 60cm2.
One area equally divided into three is 60 / 3 = 20 cm2.
As shown in Fig. 1, draw the straight line passing through the center of this figure.
As shown in Fig. 1, draw the straight line passing through the center of this figure.
A trapezoid area in a figure can be calculated as 10 × (4 +4) = 80cm2.
Thus, CP = 20 / (18 - 14) = 5cm.
Thus, CP = 20 / (18 - 14) = 5cm.
DQ = 20 / (8 - 6) =10 cm.
QR= 18 - 4/2 - 10 = 6 cm.
△BEP and △PRQ are homothetic and homothetic ratio is BE : PE = PR : QR = 2 : 6 = 1 : 3.
Thus, BE = 2 × 1/3 = 2/3.
Therefore, AB=5 - 2/3 = 13/3 cm.
Problem 12
As shown in Fig.1, there are three point A, B and P on the even ground.
There are ABCD of a 3m-high rectangular wall and PQ of a 9m-high pillar standing on the ground straight.
Fig.2 shows the picture looked at these from right above.
Wall ABCD is illuminated by the electric light in the position at the tip Q of the pillar.
Find the area of the shadow of the wall made into the ground.
Neither the size of an electric light nor the thickness of a wall shall be considered.
There are ABCD of a 3m-high rectangular wall and PQ of a 9m-high pillar standing on the ground straight.
Fig.2 shows the picture looked at these from right above.
Wall ABCD is illuminated by the electric light in the position at the tip Q of the pillar.
Find the area of the shadow of the wall made into the ground.
Neither the size of an electric light nor the thickness of a wall shall be considered.
Answer
Since △DAD' and △QPD' are homothetic in Fig.3 and as DA = 3m and QP = 9m, homothetic ratio is 3 : 9 = 1 : 3.
AP : D'P = (3 - 1) : 3 = 2 : 3.
In Fig.4, a shadow area is a shadow of wall ABCD.
Since AP : D'P = 2 : 3, △PAB and △PC'D' are homothetic.
A homothetic ratio is 2 : 3 and an area ratio is (2 × 2) : (3 × 3) = 4 : 9.
The area ratio of △PAB and a shadow area is 4 : (9 - 4) = 4 : 5.
As shown in Fig.5 as for △PAB, AF is a base and B(C) G is height.
Problem 13
45 m2
Solution
Fig.3 is the figure which looked at Fig.1 from the side.
AP : D'P = (3 - 1) : 3 = 2 : 3.
In Fig.4, a shadow area is a shadow of wall ABCD.
A homothetic ratio is 2 : 3 and an area ratio is (2 × 2) : (3 × 3) = 4 : 9.
The area ratio of △PAB and a shadow area is 4 : (9 - 4) = 4 : 5.
As shown in Fig.5 as for △PAB, AF is a base and B(C) G is height.
△BEF and △PHF are homothetic and a homothetic ratio is BE : PH = 4 : 2 = 2 : 1.
FH = 6m × 1/(2+1) = 2m.
AF = 8m + 6m - 2m = 12 m.
As EG = HP = 2m, BG = 4m + 2m = 6m.
Thus, the area of △PAB is 12 × 6 / 2 = 36 m2.
Therefore, the area of the shadow to be found is 36 m2 × 5/4= 45 m2.
Therefore, the area of the shadow to be found is 36 m2 × 5/4= 45 m2.
Problem 13
As for quadrangular pyramid O-ABCD in a figure, the bottom is a square and all the length of OA, OB, OC and OD is equal. K, L, M, N, P, Q, R, and S are the middle points of each side.
This quadrangular pyramid is cut by three planes of the plane which passes along P, K, N, R and the plane which passes along P, L, M, R and the plane which passes along S, K, L, Q and divides into some solids.
Find the ratio of the volume of a solid including the vertex O and the volume of the quadrangular pyramid O-ABCD.
This quadrangular pyramid is cut by three planes of the plane which passes along P, K, N, R and the plane which passes along P, L, M, R and the plane which passes along S, K, L, Q and divides into some solids.
Find the ratio of the volume of a solid including the vertex O and the volume of the quadrangular pyramid O-ABCD.
Answer
5 : 16
Solution
Fig. 1 is a sketch that quadrangular pyramid O-ABCD is cut by three planes written in the problem.
Fig. 2 is a sketch which is cut by the plane which passes along QMNS in addition to the three planes.
As for this quadrangular pyramid O-ABCD, it consists of six blue quadrangular pyramids and four red triangular pyramids in Fig. 2.
Each figure is extracted to Fig. 3.
When being cut by three planes, the solid including the vertex O consists of two quadrangular pyramids same as O-KLMN and one red triangular pyramid.
O-KLMN and O-ABCD are homothetic and a homothetic ratio is
OK : OA = 1 : 2.
When volume of O-ABCD is set to 1, the volume of O-KLMN is 1/2 × 1/2 × 1/2 = 1/8.
Moreover, the volume of one red triangular pyramid is/(1 - 1/8 × 6) / 4 = 1/16.
Moreover, the volume of one red triangular pyramid is/(1 - 1/8 × 6) / 4 = 1/16.
Therefore, the volume of a solid including the vertex O is 1/8 × 2 + 1/16 = 5/16 and the volume ratio to be found is 5 : 16.
