Excuse me, who has the Olympiad questions about circles

Make a regular triangle of the circumscribed circle, the point p is a point on the inferior arc ab, connect the pc to ab to d, and verify: 1 pa+1 pb=1 pd. A hint: the proband-triangle ADP is similar to the triangle PBC

The triangle is a regular triangle with four points in a circle.

So angle cpb = angle cab = 60 degrees = angle abc = angle apc angle pab = angle pcb

So the triangle APD is similar to the triangle CPB

So pc pb = pa pd 1.

PC Pa = PB PD 2 formula.

Take the point k on the PC such that pk=pb

then due to the angle cpb = 60 degrees.

So the triangle cpb is a regular triangle and we get bk=bp=pk, so the angle pkb=60 degrees.

Angle CKB = 120 degrees = Angle APB

And because of the angle bck = angle bap

Due to AAS, the triangle CKB is congruent with APB.

So ck=pa

So pc=pk+kc=pb+pa

Above 1 + 2 gets:

pc pb+pc pa=(pa+pb) pd because pa+pb=pc

So pc pb+pc pa=pc pd

Both sides are divided by pc to get :

1 Pa + 1 PB = 1 PD.

How do you circle four bottles together?

1: As shown in the figure. In a square ABCD, draw an arc with point A as the center of the circle and AB as the radius. The arc intersects with a circle of cd of diameter, O, the extension of the point, the extension of the circle, the extension of the point, and the B.C., F

1.Verify bf=cf

2.If the side length of the square is 2cm, find the length of be.

3.Verify og bc

2, but the straight line i in the center of the circle is intersected o in cd, ab is the diameter, ae is perpendicular i, perpendicular foot is perpendicular i, perpendicular foot is fVerify CE=DF

3. The central angle of the fan is 150°, and the fan area is 240, then its arc length is

If the arc length of the central angle of 100° is 5, then the radius of the circle is

Knowing that the radius of the circle at the bottom of the cone is 9 and the length of the bus bar is 90, then its surface area is

4. In the circle O, the chord AB and CD intersect at the point P, and are at equal angles to the diameter of the crossing point P (that is, the angle between the two strings and the diameter is equal), and BP=DP is verified.

5. In the circle O, the chord AB and CD intersect at the point P, and are at equal angles to the diameter of the crossing point P (that is, the angle between the two strings and the diameter is equal), and BP=DP is verified.

Hope it helps!!

I wish you success in your studies!! Happy every day!!

Suppose that before rolling, a is the left end of the bottom edge, the first time to a1, after a distance of 1 3 radius r=3 circumference, the second time to a2, after a distance of 1 3 radius r=3, and the third time to a3, is the center of rotation, the distance is 0, and returns to the initial position.

That is, after three times is a cycle, the distance is long = 2 * 3 3 + 2 * 3 + 0 = 4, after 30 rolls, that is, 10 cycles, its.

Total length of the journey = 10 * 4 = 40,

3*(2pai/3)*2*(30/3)=40pai cm

Roll 3 times for a cycle, and in each cycle, A rotates 120 degrees around B and C respectively.

The least common multiple of 50 and 30 is 150

That is, for every 150 cm run, the footprints of the two people overlap once, 150 50 = 3 for every 150 cm run, leaving a total of 3 + 5-1 = 7 footprints.

The circumference of the pool is:

157 150 = 23550 cm = m.

The diameter of the pool is: meters.

1. The sum of the areas of circle A and circle B is three-fifths of the area of circle C. The area of the shadow part in circle A accounts for one-third of the area of circle A, the area of the shadow part in circle B accounts for one-half of the area of circle B, and the area of the shadow part in circle C accounts for one-quarter of the area of circle C.

Let the area of circle A be x, let the area of circle B be y, and let the area of circle C be z.

x+y=3z/5

x/3+y/2=z/4

The solution yields x=3z 10 and y=3z 10

x:y=1:1

Solution: From the meaning of the question, it can be seen that the quadrilateral OAC is a prism, and the triangle OAB and the triangle OAC are both equilateral triangles, then OA bisects BC perpendicularly, and BC=6 3 can be obtained according to the Pythagorean theorem of the RT triangle

Solution: Circle through (-1,2)(7,2) two points.

The center of the circle is on the perpendicular bisector of these two points, i.e. the center of the circle is on the straight line x=3.

Let the center coordinate of the circle be (3,m).

The square of the distance from the center of the circle to (-1,2) is (3+1) +m-2) =16+(m-2).

and the distance from the center of the circle to (3,10) is |10-m|The square of the distance is (10-m).

16+(m-2)²=10-m)²

m=5 The coordinates of the center of the circle are (3,5) and the radius is |10-5|=5 The distance from the center of the circle to the x-axis is 5

The circle is tangent to the x-axis.

distancing. The y of coordinates (-1,2) (7,2) (3,10) is greater than 0 and the coordinates (-1,2) (7,2) (3,10) are all above the x-axis.

The circle passes through these three points, indicating that the circle is above the x-axis.

The circle is separated from the x-axis.

Can the first floor be written? It's still separate.

Substituting these three points into the general equation of a circle: x 2 + y 2 + dx + ey + f = 0 gives :

5-d+2e+f=0

53+7d+2e+f=0

109+3d+10e+f=0

The solution is: d=-6, e=-10, f=9

The equation for the circle that can be turned is: (x-3) 2 + y-5) 2=25=5 2 The circle is tangent to the x-axis.

The intersecting string theorem is mainly used.

1) Rectangle.

According to the intersecting string theorem, we get:

ad^2=df*de

bc^2=ce*cf

And because ad=bc

So df*de=ce*cf where de=df+ef, cf=ce+ef, so (df+ef)*df=ce*(ce+ef) where ef=ef, therefore, in order for the equation to hold, then df=ce(2) isosceles trapezoidal (the same principle).

According to the intersecting string theorem, we get:

ad^2=df*de

bc^2=ce*cf

And because AD=BC (isosceles trapezoidal, waist equal).

So df*de=ce*cf where de=df+ef, cf=ce+ef, so (df+ef)*df=ce*(ce+ef) where ef=ef, therefore, to make the equation true, then df=ce(3) right-angled trapezoidal (waist unequal).

According to the intersecting string theorem, we get:

ad^2=df*de

bc^2=ce*cf

where AD is not equal to BC, DE=DF+EF, CF=CE+EF, so. ad^2=(df+ef)*df

bc^2=(ce+ef)*ce

where EF=EF, therefore, if you want to make DF=CF, you need AD=BC, and because the quadrilateral is a right-angled trapezoid, AB is not equal to BC, so DF is not equal to CF In summary, if you want to make DF=CF, you need AD=BC

The above is done according to your requirements, in fact, it doesn't have to be so complicated, the following ones will do.

According to the intersecting string theorem, we get:

ad^2=df*de

bc^2=ce*cf

where de=df+ef, cf=ce+ef

So. ad^2=(df+ef)*df

bc^2=(ce+ef)*ce

where EF=EF, so if you want to make DF=CF, you need AD=BC, so when AD=BC, DF=CF

According to the intersecting string theorem, we get:

ad^2=df*de

bc^2=ce*cf

where de=df+ef, cf=ce+ef

So. ad^2=(df+ef)*df

bc^2=(ce+ef)*ce

where EF=EF, so if you want to make DF=CF, you need AD=BC, so when AD=BC, DF=CF

PEF is an isosceles right triangle.

Proof: Connect PA and PB

AB is the diameter, AQB EQF 90°

EF is the diameter of O and EPF 90° in APE and BPF.

pa＝pb，pbf＝∠pae

ape=∠bpf=90°+∠epb，△ape≌△bpf

PE=PF, PEF is an isosceles right triangle.

EPF is a right triangle.

Since AB is the diameter, the angle AQB is a right angle.

So the angle EQF is also a right angle, so EF is O'diameter.

So EFP is also a right triangle. Proof is complete.

Proof: Connect PA, PB, PQ

ab is the diameter , apb = aqb eqf 90°

epf＝∠eqf=90°

p is the midpoint of the arc AB.

pba＝∠pab=45°

pqa=∠pba=45°

pfe=∠pqa=45°

PEF is an isosceles right triangle.