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.
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
When the circles P and Q come to the angle of the vertex B, it looks as it is shown in Fig. 3.
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.
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.
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 x 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.
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.
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 x 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.
