Time : 60 minutes
Passing mark : 70 %
Answer : End of the problem
Problem 1
The quadrangle ABCD of the figure is a parallelogram. E, P and G are points on the sides AB, CD and DA, respectively and AE = EB and AG=GD and DF : FC = 4 : 1.
Moreover, the point H is an intersection of EF and BG.
The point I is an intersection of EF and CG.
Answer the following questions.
(1) The intersection of the straight line which extended side DA and the straight line which extended FE is set to P.
Moreover, the intersection of the straight line which extended the side BC and the straight line which extended EF is set to Q.
In this case, Find PA : AD and BC : CQ.
(2) Find GH : HB and GI : IC.
(3) Find the ratio of the area of the triangle AHG, the triangle GHI and the triangle GID.
Problem 2
Problem 3
There are three integers A, B and C which are A + B + C = 1000.
B/A is calculated to the first decimal place and it is 7 when rounding off the first decimal place of the quotient.
Moreover, when C is divided by B, the quotient is 2 and remainder is 16.
Answer the following questions.
(1) When B/A is calculated, it can be divided at the first decimal place exactly and the quotient is 6.5.
In this case, find A, B and C, respectively.
(2) In other than (1), find all the numbers considered as a combination of A, B and C.
Problem 4
The quadrangle ABCD of the figure is a parallelogram. E, P and G are points on the sides AB, CD and DA, respectively and AE = EB and AG=GD and DF : FC = 4 : 1.
Moreover, the point H is an intersection of EF and BG.
The point I is an intersection of EF and CG.
Answer the following questions.
(1) The intersection of the straight line which extended side DA and the straight line which extended FE is set to P.
Moreover, the intersection of the straight line which extended the side BC and the straight line which extended EF is set to Q.
In this case, Find PA : AD and BC : CQ.
(2) Find GH : HB and GI : IC.
(3) Find the ratio of the area of the triangle AHG, the triangle GHI and the triangle GID.
Problem 2
Answer the following questions about the hour hand (short hand), the minute hand (long hand) and the second hand of the clock in 60 minutes from 10:10 to 11:10.
(1) Find the time when an hour hand and the minute hand overlap.
The second only should answer using a fraction.
(2) Considering the case that any two of three hands do not have overlapped either among three hands, an hour hand, the minute hand and the second hand.
In this case, the center of the clock is divided into three angles as shown in a figure.
Find the number of times that two angles become equal among these from the time when it is found at (1) until 11:10 and also find the last time when two angles become equal.
As for the time, the second only should answer using a fraction.
(1) Find the time when an hour hand and the minute hand overlap.
The second only should answer using a fraction.
(2) Considering the case that any two of three hands do not have overlapped either among three hands, an hour hand, the minute hand and the second hand.
In this case, the center of the clock is divided into three angles as shown in a figure.
Find the number of times that two angles become equal among these from the time when it is found at (1) until 11:10 and also find the last time when two angles become equal.
As for the time, the second only should answer using a fraction.
Problem 3
B/A is calculated to the first decimal place and it is 7 when rounding off the first decimal place of the quotient.
Moreover, when C is divided by B, the quotient is 2 and remainder is 16.
Answer the following questions.
(1) When B/A is calculated, it can be divided at the first decimal place exactly and the quotient is 6.5.
In this case, find A, B and C, respectively.
(2) In other than (1), find all the numbers considered as a combination of A, B and C.
Problem 4
N sheets of card in which 1, 2, 3, ...., N is written each are clockwise arranged in the small order of the number circularly.
When removing one card at a time in accordance with the following rule, it is considered which card remains at the end.
<Rule 1>
<Rule 1>
The card in which 1 is written is removed first.
<Rule 2>
<Rule 2>
When a certain card is removed, the 2nd card from the card counted clockwise and will be removed.
This operation is repeated until only one card remains.
For example, in case of N = 13, cards are removed as shown in Fig. 1, and #10 card finally remains.
(× mark expresses the removed card.)
Answer the following questions.
(1) Answer the number written to the card which remains at the end in case of N = 8.
(2) When cards are removed at the 1st round in case of N = 16, as shown in Fig. 2, eight cards remain.
The card in which 2 was written is removed next.
Answer the number written to the card which remains at the end in case of N = 16.
Moreover, answer the number written to the card which remains at the end, respectively in case of N = 32 and N = 64.
(3) When the cards in which 1, 3, and 5 are written at the 1st round are removed in case of N = 35, 32 cards remain.
The card in which 7 was written is removed next.
Answer the number written to the card which remains at the end in case of N = 35.
(4) When 36 cards are removed to the 1st round in case of N = 100, 64 cards remain.
Answer the number written to the card which remains at the end in case of N = 100.
(5) Answer the number written to the card which remains at the end in case of N = 2009.
<Answer>
Problem 1The quadrangle ABCD of the figure is a parallelogram. E, P and G are points on the sides AB, CD and DA, respectively and AE = EB and AG=GD and DF : FC = 4 : 1.
Moreover, the point H is an intersection of EF and BG.
The point I is an intersection of EF and CG.
Answer the following questions.
(1) The intersection of the straight line which extended side DA and the straight line which extended FE is set to P.
Moreover, the intersection of the straight line which extended the side BC and the straight line which extended EF is set to Q.
In this case, Find PA : AD and BC : CQ.
(2) Find GH : HB and GI : IC.
(3) Find the ratio of the area of the triangle AHG, the triangle GHI and the triangle GID.
Problem 2
Problem 3
Problem 4
Moreover, the point H is an intersection of EF and BG.
The point I is an intersection of EF and CG.
Answer the following questions.
(1) The intersection of the straight line which extended side DA and the straight line which extended FE is set to P.
Moreover, the intersection of the straight line which extended the side BC and the straight line which extended EF is set to Q.
In this case, Find PA : AD and BC : CQ.
(2) Find GH : HB and GI : IC.
(3) Find the ratio of the area of the triangle AHG, the triangle GHI and the triangle GID.
Answer
(1) PA : AD = 5 : 3, BC : CQ = 3 : 2
(2) GH : HB = 13 : 10, GI : IC = 13 : 4
(3) 17 : 26 : 23
Solution
(1)
AE : DF = 1/2 : 4/5 = 5 : 8.
△PAE∽△PDF
Homothetic ratio : △PAE : △PDF = AE : DF = 5 : 8.
As PA : PD = 5 : 8, PA : AD = 5 : (8 - 5) = 5 : 3.
BE : CF = 1/2 : 1/5 = 5 : 2.
△QBE∽△QCF
Homothetic ratio : △QBE : △QCF = BE : CF = 5 : 2.
As BQ : CQ = 5 : 2, BC : CQ = (5 - 2) : 2 = 3 : 2.
(1) PA : AD = 5 : 3, BC : CQ = 3 : 2
(2) GH : HB = 13 : 10, GI : IC = 13 : 4
(3) 17 : 26 : 23
Solution
(1)
AE : DF = 1/2 : 4/5 = 5 : 8.
△PAE∽△PDF
Homothetic ratio : △PAE : △PDF = AE : DF = 5 : 8.
As PA : PD = 5 : 8, PA : AD = 5 : (8 - 5) = 5 : 3.
BE : CF = 1/2 : 1/5 = 5 : 2.
△QBE∽△QCF
Homothetic ratio : △QBE : △QCF = BE : CF = 5 : 2.
As BQ : CQ = 5 : 2, BC : CQ = (5 - 2) : 2 = 3 : 2.
(2)
The length of AD is set to 6, PA = 6 × 5/3 = 10 and PG = 10 + 3 = 13.
As BC = AD = 6, CQ = 6 × 2/3 = 4 and QB = 4 + 6 = 10.
△PGH∽△QBH
Homothetic ratio : △PGH : △QBH = PG : QB = 13 : 10.
Thus GH : HB is also 13 : 10.
△PGI∽△QCI
Homothetic ratio : △PGI : △QCI = PG : QC = 13 : 4.
Thus GI : IC is also 13 : 4.
(3)
The area of quadrangle ABCD is set to be 1.
The area of △ABG = 1 × 1/4 = 1/4.
Then the area of △AHG = 1/4 × 13/(13+10) = 13/(4×23).
△GBC = 1 × 1/2 = 1/2.
△GHI = △GBC × 13/(13+10) × 13/(13+4) = 1/2 × 13/23 × 13/17 = 169/(2×23×17)
△DCG = 1 × 1/4 = 1/4.
△GID = 1/4 × 13/(13+4) = 13/(4×17).
△AHG : △GHI : △GID = 13/(4×23) : 169/(2×23×17) : 13/(4×17) = 17×13 : 13×13×2 : 13×23/ (2×23×17) = 17 : 26 : 23.
Problem 2
Answer the following questions about the hour hand (short hand), the minute hand (long hand) and the second hand of the clock in 60 minutes from 10:10 to 11:10.
(1) Find the time when an hour hand and the minute hand overlap.
The second only should answer using a fraction.
(2) Considering the case that any two of three hands do not have overlapped either among three hands, an hour hand, the minute hand and the second hand.
In this case, the center of the clock is divided into three angles as shown in a figure.
Find the number of times that two angles become equal among these from the time when it is found at (1) until 11:10 and also find the last time when two angles become equal.
As for the time, the second only should answer using a fraction.
Thus, the number of times that two angles become equal for 15 minutes from 10:54 360/11 seconds to 11:09 360/11 seconds is 4 times × 15= 60 times.
(1) Find the time when an hour hand and the minute hand overlap.
The second only should answer using a fraction.
(2) Considering the case that any two of three hands do not have overlapped either among three hands, an hour hand, the minute hand and the second hand.
In this case, the center of the clock is divided into three angles as shown in a figure.
Find the number of times that two angles become equal among these from the time when it is found at (1) until 11:10 and also find the last time when two angles become equal.
As for the time, the second only should answer using a fraction.
Answer
(1) 10:54 360/11 seconds
(2) 61 times
11:09 3060/73 seconds
(2) 61 times
11:09 3060/73 seconds
Solution
(1)
At 10:00, the difference of angle between hour hand and minute hand is 300 degrees as shown below.
Minute hand moves 6 degrees per minute and hour hand moves 0.5 degrees per minute.
The time for the minute hand to overlap the hour hand is 300 / (6 - 0.5) = 600/11 = 54 and 6/11 minutes.
6/11 minutes = 60 seconds × 6/11 = 360/11 seconds.
(2)
It turns out that in 1 minute while the second hand rotates one round from 10:54 360/11 seconds, there are four times that two angles become equal among three angles as shown in Fig. 2.
Furthermore, since there is 1 time in from 11:09 360/11 seconds to 11:10 as shown in Fig. 3, it is 60 + 1 = 61 times in all.
It is the time of Fig. 3 when two angles become equal at the last time and the time is considered from 11:09.
The angle of second hand and hour hand at 11:09 is 30 degrees × 11 + 0.5 degrees × 9 minutes =334.5 degrees (Angle of the red of Fig. 4).
In 1 second, the second hand moves 360 degrees / 60 second = 6 degrees and the hour hand moves 0.5 degree / 60 seconds =5/600= 1/120 degrees.
The angle of the second hand and the hour hand in 1 second becomes small by 6 - 1/120 = 719/120 degrees.
The angle of hour hand and minute hand at 11:09 is 30 degrees + (6 - 0.5) × 9 minutes = 79.5 degrees (blue angle of Fig. 4).
In 1 second, the minute hand moves 6 degrees / 60 second = 1/10 degrees.
The angle of the hour hand and the minute hand in 1 second becomes large by 1/10 - 1/120 = 11/120 degrees.
The time for two angles to be equal is (334.5 - 79.5) / (719/120 + 11/120) = 255 / 73/12 = 3060/73 seconds.
Therefore the time to be found is 11:09 3060/73 seconds.
Problem 3
There are three integers A, B and C which are A + B + C = 1000.
B/A is calculated to the first decimal place and it is 7 when rounding off the first decimal place of the quotient.
Moreover, when C is divided by B, the quotient is 2 and remainder is 16.
Answer the following questions.
(1) When B/A is calculated, it can be divided at the first decimal place exactly and the quotient is 6.5.
In this case, find A, B and C, respectively.
(2) In other than (1), find all the numbers considered as a combination of A, B and C.
B/A is calculated to the first decimal place and it is 7 when rounding off the first decimal place of the quotient.
Moreover, when C is divided by B, the quotient is 2 and remainder is 16.
Answer the following questions.
(1) When B/A is calculated, it can be divided at the first decimal place exactly and the quotient is 6.5.
In this case, find A, B and C, respectively.
(2) In other than (1), find all the numbers considered as a combination of A, B and C.
Answer
(1) (A,B,C) = (48, 312, 640)
(2) (A,B,C) = (42, 314, 644) = (45, 313, 642)
(2) (A,B,C) = (42, 314, 644) = (45, 313, 642)
Solution
(1)
As B / A = 6.5, B = A × 6.5.
As C / B = 2, remainder 16, C = B × 2 + 16 = (A × 6.5) × 2 + 16 = A × 13 + 16.
A + B + C = A + A × 6.5 + A × 13 + 16 = A × 20.5 + 16 = 1000.
A = (1000 - 16 ) / 20.5 = 48.
B = 48 × 6.5 = 312.
C = 48 × 13 + 16 = 640.
(2)
In case B / A = 7.5, B = A × 7.5.
C = (A × 7.5) × 2 + 16 = A × 15 + 16.
A + B + C = A + A × 7.5 + A × 15 + 16 = A × 23.5 + 16 = 1000.
A = (1000 - 16 ) / 23.5 = 41.----.
Then the range of A is between 42 and 48.
A + B + C = A + B + B × 2 + 16 = 1000
A + B × 3 = 984.
Since B × 3 and 984 are multiples of 3, A is also multiple of 3.
Multiples of 3 between 42 and 48 are 42, 45, 48.
When A = 42, B = (984 - 42) / 3 = 314, C = 314 × 2 + 16 = 644.
When A = 45, B = (984 - 45) / 3 = 313, C = 313 × 2 + 16 = 642.
Problem 4
N sheets of card in which 1, 2, 3, ...., N is written each are clockwise arranged in the small order of the number circularly.
When removing one card at a time in accordance with the following rule, it is considered which card remains at the end.
<Rule 1>
<Rule 1>
The card in which 1 is written is removed first.
<Rule 2>
<Rule 2>
When a certain card is removed, the 2nd card from the card counted clockwise and will be removed.
This operation is repeated until only one card remains.
For example, in case of N = 13, cards are removed as shown in Fig. 1, and #10 card finally remains.
(× mark expresses the removed card.)
Answer the following questions.
(1) Answer the number written to the card which remains at the end in case of N = 8.
(2) When cards are removed at the 1st round in case of N = 16, as shown in Fig. 2, eight cards remain.
The card in which 2 was written is removed next.
Answer the number written to the card which remains at the end in case of N = 16.
Moreover, answer the number written to the card which remains at the end, respectively in case of N = 32 and N = 64.
(3) When the cards in which 1, 3, and 5 are written at the 1st round are removed in case of N = 35, 32 cards remain.
The card in which 7 was written is removed next.
Answer the number written to the card which remains at the end in case of N = 35.
(4) When 36 cards are removed to the 1st round in case of N = 100, 64 cards remain.
Answer the number written to the card which remains at the end in case of N = 100.
(5) Answer the number written to the card which remains at the end in case of N = 2009.
Answer
(1) 8
(2) 16, 32, 64
(3) 6
(4) 72
(5) 1970
Solution
(1) Arrange cards from 1 to 8 and remove.
Cards being removed at the 1st round is 1, 3, 5, 7.
At the 2nd round is 2, 6.
At the 3rd week is 4.
Therefore, the remaining card is 8.
(2) Remove cards using Fig. 2.
Cards being removed at the 1st round is 1, 3, 5, 7, 9, 11, 13, 15.
At the 2nd round is 2, 6, 10, 14.
At the 3rd round is 4, 12.
At the 4th round is 8.
Therefore, the remaining card is 16.
As for the case of n= 32, odd number is removed at the 1st round.
Since it begins from 2 and card is removed every four at the 2nd round, cards removed is 2, 6, 10, 14, 18, 22, 26, 30.
Since it begins from 4 and card is removed every eight at the 3rd week, cards removed is 4, 12, 20, 28.
Since it begins from 8 and card is removed every 16 at the 4th week, cards removed is 8, 24.
At the 5th round is 16.
Therefore, the remaining card is 32.
This operation shows that in the case of n = 2 × 2 × 2 = 8, n = 2 × 2 × 2 × 2 = 16, and n = 2 × 2 × 2 × 2 × 2 = 32, 8, 16, and 32 remain.
The card in which it remains in the case of 64= 2 × 2 × 2 × 2 × 2 × 2 is 64.
(3) In case of n = 35, when three cards, 1, 3, and 5, are removed, it can be considered as n= 32.
And in the case of n= 32, it began to take from 1, but in this case, it can be considered that it begins to take from 7 which is next to 5.
In the case of n= 32, it began to take from 1 and the remaining number is 32 which is just before 1 in the arrangement of numbers.
When it thinks the same way also in this case, the number to be remained is 6 which is just before 7 which is the number to be removed first.
(4) Consider as the same way as (3).
Since the 37th card removed is the 37th odd number counted from 1, it becomes 1, 3, 5, 7, 9 -------, 1 + 2 × (37 - 1) = 73.
Therefore, the card which remains in this case is 72 which is just before 73 which is considered as the first card removed.
(5) 64 × 2 = 128, 128 × 2 = 256, 256 × 2 = 512, 512 × 2 = 1024, 1024 × 2 = 2048.
Among these numbers, the number which is less than 2009 and is nearest to 2009 is 1024.
According to the calculation of 2009 - 1024 = 985, when 985 pieces are removed, it can be considered as the case of 1024.
The 985th odd number is 1 + 2 × (985 - 1) = 1969.
1024 pieces of cards remain when 1969 is removed.
Therefore, since the first card to be removed is 1971, the card to remain is 1970 which is just before 1971.
