Time : 60 minutes
Passing marks : 70%
Passing marks : 70%
Answer : End of the problem
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Problem 1
Problem 3
Problem 4
Problem 4
I divide integers more than 6 by 6 and add a remainder to a quotient.
I express this calculation in → and continue it until it becomes 5 or less.
<Example 1>
12 / 6 = 2 remainder 0, then 2 + 0 = 2.
When begin with 12, it is expressed as 12 → 2.
<Example 2>
50 / 6 = 8 remainder 2, then 8 + 2 = 10, 10 / 6 = 1 remainder 4, then 1 + 4 = 5.
When begin with 50, it is expressed as 50 → 10 → 5.
Answer the following questions.
(1) When I begin this calculation with 2003, express the result by using →.
(2) Find the number of integers more than 6 to be suitable for A when it is expressed in one arrow as A → B.
(3) Find the smallest integer in integers more than 6 to be suitable for A when it is expressed in two arrows as A → B → C.
In addition, find the biggest integer.
(4) Find the smallest integer in integers more than 6 to be suitable for A when it is expressed in four arrows as A → B → C → D → E.
In addition, find the biggest integer.
Problem 2
A rectangle departs from the position in a figure and is moving in the direction of an arrow at 2 cm/s.
(1) Find the time when the rectangle is completely inside of the triangle.
(2) Find the area of the portion of the rectangle out of a triangle 12 seconds after leaving.
(3) Find the time when the area of the portion of the rectangle which is contained in the triangle is 30 cm2 or more.
Problem 3
The figure shows a cube whose length of one side is 6cm.
Two points of P,Q moves on the side of the cube with a speed of 2cm per second, 3cm per second respectively.
P,Q leave A at the same time.
P moves in order of A → E → H → G → C → B → A and repeats this cycle.
Q moves in order of A → D → C → G → F → E → A and repeats this cycle.
Answer the following questions.
(1) Find the time when P and Q meet for the first time after leaving A.
(2) Find the time when P and Q meet for the fifth time after leaving A.
Problem 4
There is a flower clock in a certain town and the long hand advances one round in one hour with a fixed speed.
The hour hand stands still first 59 seconds in every minute and advances 1 of 720 round with a fixed speed in last one second.
The long hand and the hour hand have overlapped exactly at 0:00 a.m.
Answer to the following questions.
(1) Find the time when the long hand and the hour hand become right-angled during 0:00 a.m. and 1:00 a.m.
(2) Find the time when the long hand and the hour hand become right-angled during 8:00 a.m. and 9:00 a.m. except exactly at 9:00 a.m.
Problem 5
(2) C and D were connected as shown in Fig.2.
Find the area ratio of the sum of the area of four triangles painted black and the area of one parallelogram.
The hour hand stands still first 59 seconds in every minute and advances 1 of 720 round with a fixed speed in last one second.
The long hand and the hour hand have overlapped exactly at 0:00 a.m.
Answer to the following questions.
(1) Find the time when the long hand and the hour hand become right-angled during 0:00 a.m. and 1:00 a.m.
(2) Find the time when the long hand and the hour hand become right-angled during 8:00 a.m. and 9:00 a.m. except exactly at 9:00 a.m.
Problem 5
There are some parallelograms whose length of two sides is 2 cm and 1 cm, and one inside angle is 60 degrees.
These are connected at the vertex so that all of the 2 cm sides are parallel as shown in Fig.1 and Fig.2.
(1) A and B were connected as shown in Fig.1.
Find the area ratio of the sum of the area of two triangles painted black and the area of one parallelogram.
These are connected at the vertex so that all of the 2 cm sides are parallel as shown in Fig.1 and Fig.2.
(1) A and B were connected as shown in Fig.1.
Find the area ratio of the sum of the area of two triangles painted black and the area of one parallelogram.
Find the area ratio of the sum of the area of four triangles painted black and the area of one parallelogram.
<Answer>
Problem 1
I divide integers more than 6 by 6 and add a remainder to a quotient.
I express this calculation in → and continue it until it becomes 5 or less.
<Example 1>
12 / 6 = 2 remainder 0, then 2 + 0 = 2.
When begin with 12, it is expressed as 12 → 2.
<Example 2)
50 / 6 = 8 remainder 2, then 8 + 2 = 10, 10 / 6 = 1 remainder 4, then 1 + 4 = 5.
When begin with 50, it is expressed as 50 → 10 → 5.
Answer the following questions.
(1) When I begin this calculation with 2003, express the result by using →.
(2) Find the number of integers more than 6 to be suitable for A when it is expressed in one arrow as A → B.
(3) Find the smallest integer in integers more than 6 to be suitable for A when it is expressed in two arrows as A → B → C.
In addition, find the biggest integer.
(4) Find the smallest integer in integers more than 6 to be suitable for A when it is expressed in four arrows as A → B → C → D → E.
In addition, find the biggest integer.
Answer
(1) 2003 → 338 → 58 → 13 → 3
(2) 15 pieces
(3) 11, 180
(4) 221, 6480
Problem 2
A rectangle departs from the position in a figure and is moving in the direction of an arrow at 2 cm/s.
(1) Find the time when the rectangle is completely inside of the triangle.
(2) Find the area of the portion of the rectangle out of a triangle 12 seconds after leaving.
(3) Find the time when the area of the portion of the rectangle which is contained in the triangle is 30 cm2 or more.
Answer
(2) In 12 seconds, a rectangle moves 2 cm × 12= 24 cm.
(3) When a part of area of the rectangle in a triangle is 30 cm2, the sum of the length of a upper base and a lower base is 30 × 2 / 6 = 10cm.
Problem 3
(1) 1.5 seconds
(2) 55.5 cm2
(3) 9.75 seconds
(2) 55.5 cm2
(3) 9.75 seconds
Solution
(1) While the rectangle is moving the length of the arrow in Fig.1, it is completely inside of the triangle.
In order to find this length, the length of DE is to be found.
△ABC and △ADE are homothetic and the ratio of height is 24 : (24 - 6) = 24 : 18 = 4 : 3.
Since BC : DE = 4 : 3, the length of DE is 20 cm × 3/4= 15 cm.
Thus, the length of the arrow is 15 - 12 =3 cm.
The time to be found is 3 / 2 = 1.5 seconds since a rectangle moves at 2 cm/s.
(2) In 12 seconds, a rectangle moves 2 cm × 12= 24 cm.
Fig. 2 is a figure of 12 seconds after.
BF = 24 - 20 = 4 cm.
The area to find is the area which trapezoid DBFG area which is the inside of triangle is subtracted from the rectangular area.
In Fig. 2, △ABI and △HBF are homothetic and as BI : AI =10: 24 = 5 : 12, BF : HF = 5 : 12.
HF = 4 cm × 12/5 = 48/5 cm.
HG=48/5 - 6 = 18/5 cm.
Since △HBF and △HDG are also homothetic, DG = HG × 5/12 = 18/5 × 5/12 = 1.5cm.
Therefore, since the area of trapezoid DBFG is (1.5 + 4) × 6 / 2 = 16.5 cm2, the area to be found is 6 × 12 - 16.5 = 55.5cm2.
(3) When a part of area of the rectangle in a triangle is 30 cm2, the sum of the length of a upper base and a lower base is 30 × 2 / 6 = 10cm.
According to (2), the difference of the length of DG and BF is 4 - 1.5 = 2.5 cm.
This difference of length does not change while the form where a rectangle and a triangle overlap is a trapezoid.
A part of area of the rectangle in a triangle becomes 30 cm2 for the first time is at the time of BF= (10 + 2 - 5) / 2 = 6.25cm.
It is again becomes 30 cm2 is at the time of QC = 6.25cm as shown in Fig. 3.
That is, the time to find is the time that the rectangle moved the length of the arrow in Fig. 3.
The length is 12 + (20 - 6.25 × 2) = 19.5 cm. Time to move this length is for 19.5 / 2 = 9.75 seconds.
Problem 3
The figure shows a cube whose length of one side is 6cm.
Two points of P,Q moves on the side of the cube with a speed of 2cm per second, 3cm per second respectively.
P,Q leave A at the same time.
P moves in order of A → E → H → G → C → B → A and repeats this cycle.
Q moves in order of A → D → C → G → F → E → A and repeats this cycle.
Answer the following questions.
(1) Find the time when P and Q meet for the first time after leaving A.
(2) Find the time when P and Q meet for the fifth time after leaving A.
Answer
(1) 28.8 seconds
(2) 100.8 seconds
Problem 4
There is a flower clock in a certain town and the long hand advances one round in one hour with a fixed speed.
Problem 5
(2) C and D were connected as shown in Fig.2.
Find the area ratio of the sum of the area of four triangles painted black and the area of one parallelogram.
The hour hand stands still first 59 seconds in every minute and advances 1 of 720 round with a fixed speed in last one second.
The long hand and the hour hand have overlapped exactly at 0:00 a.m.
The long hand and the hour hand have overlapped exactly at 0:00 a.m.
Answer to the following questions.
(1) Find the time when the long hand and the hour hand become right-angled during 0:00 a.m. and 1:00 a.m.
(2) Find the time when the long hand and the hour hand become right-angled during 8:00 a.m. and 9:00 a.m. except exactly at 9:00 a.m.
Answer
(1) 0:16:20 a.m. and 0:49:05 a.m.
(2) 8:27:15 a.m. and 8:59:55 a.m.
(2) 8:27:15 a.m. and 8:59:55 a.m.
Solution
(1) 1st time : The long hand advances 6 degrees in 1 minute (360 degrees / 60 minutes) and the hour hand advances 0.5 degrees in 1 minute (360 degrees × 1/720).
In order to find the time making the the difference of 90 degrees between the long hand and the hour hand, 90 degrees / ( 6 degrees - 0.5 degrees) = 16.33----- minutes.
This indicates that the angle between the long hand and the hour hand at 0:16 00 is 88 degrees ((6 degrees - 0.5 degrees) × 16 minutes = 88).
This indicates that the angle between the long hand and the hour hand at 0:16 00 is 88 degrees ((6 degrees - 0.5 degrees) × 16 minutes = 88).
As for this clock, during the following 59 seconds, only the long hand advances.
In order for the angle between the long hand and the hour hand to be wider by 2 degrees more (90 - 88 = 2), it takes 20 seconds (2 degrees / 6 degrees = 1/3 minute =20 seconds).
Therefore, the time to be found is 0:16:20 a.m.
2nd time : When the long hand advances more than the hour hand by 270 degrees (360 - 90 = 270), the angle is to become 90 degrees again.
The time is calculated as 270 degree / (6 degrees - 0.5 degrees) = 49.0--- minutes.
This indicates that the angle between the long hand and the hour hand at 0:49 00 is 90.5 degrees
The time is calculated as 270 degree / (6 degrees - 0.5 degrees) = 49.0--- minutes.
This indicates that the angle between the long hand and the hour hand at 0:49 00 is 90.5 degrees
360 degrees - (6 degrees - 0.5 degrees) × 49 minutes = 90.5.
As for this clock, during the following 59 seconds, only the long hand advances.
In order for the angle between the long hand and the hour hand to be narrower by 0.5 degrees (90.5 - 90 = 0.5), it takes 5 seconds
0.5 degrees / 6 degrees = 1/12 minute = 5 seconds.
Therefore, the time to be found is 0:49:05 a.m.
(2) The way of thinking is the same as (1).
1st time : The angles between the long hand and the hour hand is 30 degrees x 8= 240 degrees at 8:00.
1st time : The angles between the long hand and the hour hand is 30 degrees x 8= 240 degrees at 8:00.
According to the calculation of (240 - 90) / (6 - 0.5) = 27.2 ... , the angle at 8:27:00 is 240 - (6 - 0.5) × 27 = 91.5 degrees.
During the following 59 seconds, only the long hand advances.
During the following 59 seconds, only the long hand advances.
In order for the angle between the long hand and the hour hand to be narrower by 1.5 degrees (91.5 - 90 = 1.5), it takes 15 seconds (1.5 degrees / 6 degrees = 1/4 minute = 15 seconds).
Therefore, the time to be found is 8:27:15 a.m.
2nd time : The angles between long and hour is (6 - 0.5) × 59 = 84.5 degrees at 8:59.
During the following 59 seconds, only the long hand advances.
During the following 59 seconds, only the long hand advances.
In order for the angle between the long hand and the hour hand to be wider by 5.5 degrees (90 - 84.5 = 5.5), it takes 55 seconds
5.5 degrees / 6 degrees = 55/60 minute = 55 seconds.
Therefore, the time to be found is 8:59:55 a.m.
Note : During 59 seconds after 8:59 only the long hand advances.
The angle is 90 degrees at 8:59:55 and it will be 90 degrees again exactly at 9:00.
Problem 5
There are some parallelograms whose length of two sides is 2 cm and 1 cm, and one inside angle is 60 degrees.
These are connected at the vertex so that all of the 2 cm sides are parallel as shown in Fig.1 and Fig.2.
(1) A and B were connected as shown in Fig.1.
Find the area ratio of the sum of the area of two triangles painted black and the area of one parallelogram.
These are connected at the vertex so that all of the 2 cm sides are parallel as shown in Fig.1 and Fig.2.
(1) A and B were connected as shown in Fig.1.
Find the area ratio of the sum of the area of two triangles painted black and the area of one parallelogram.
Find the area ratio of the sum of the area of four triangles painted black and the area of one parallelogram.
Answer
(1) 1 : 42
(2) 1 : 24
(2) 1 : 24
Solution
（1）In Fig.1 the portion divided up and down by line AB is the same form and two triangles painted black are congruent.
△ACB is created by drawing an auxiliary line as shown in Fig.3.
A red triangle turns into an equilateral triangle whose one side is 1 cm from the conditions in question.
A red triangle turns into an equilateral triangle whose one side is 1 cm from the conditions in question.
The length of CB is 2cm + 1cm + 2cm + 2cm = 7cm.
△ACB and △ADG are homothetic and a homothetic ratio of AC : AD = 3 : 1.
Therefore, DG = CB × 1/3 = 7 × 1/3 = 7/3cm.
Since DE = 2cm, EG = 7/3 - 2 = 1/3cm.
△AHF and △EGH are homothetic and a homothetic ratio is AF : EG = 2 : 1/3 = 6 : 1.
An area ratio is 6 × 6 : 1 × 1 = 36 : 1.
The area ratio of △AHF of △EGH is 1 : 36.
In Fig.4 the area of △AEF is 1/2 times that of the parallelogram and △AHF is the 6 / (6+1) = 6/7 times of △AEF.
△AHF is the 1/2 × 6/7 = 3/7 times of the parallelogram.
Therefore, △EGH is the 3/7 × 1/36 = 1/84 time of the parallelogram.
The area ratio of a black area and the parallelogram is 1/84 × 2 : 1 = 1 : 42.
(2) In Fig.5 ED = 2 + 1 + 2 + 2 + 1 + 2 + 2 = 12cm.
△CED and △CFI are homothetic and a homothetic ratio is 5 : 1.
FI = ED × 1/5 = 12 × 1/5 = 12/5cm.
Thus, GI = 12/5 - 2 = 2/5cm.
A homothetic ratio of △CJH and △JGI are 2/5 : 2 = 1 : 5 and an area ratio is 1 : 25.
△CJH is 1/2 × 5 / (1+5) = 5/12 of the area of parallelogram.
△JGI is 5/12 × 1/25 = 1/60 of the area of parallelogram. ----<1>
Next, since the homothetic ratio of △CED and △CKL is 5 : 2, KL = 12 × 2/5 = 24/5cm.
Thus, LN= KN - KL = (2 + 1 + 2) - 24/5 = 1/5cm.
△JGI and △LMN are homothetic and homothetic ratio is GI : LN = 2/5 : 1/5 = 2 : 1 and an area ratio is 4 : 1.
According to <1>, the area of △LMN 1/60 × 1/4 = 1/240 of the area of a parallelogram----<2>
The area of a black triangle is △JGI × 2 + △LMN × 2.
<1> × 2 + <2> × 2 = 1/60 × 2 + 1/240 × 2 =1/24 time.
The area ratio of the sum of black triangles and a parallelogram is 1/24 :1 = 1: 24.
