High score bounty Analytic geometry problems. Ask the master to write out the process of solving the

Dude, do you want a detailed answer process? Or do you want the answer directly? Forget it, I've uploaded the detailed process with a diagram, so you can take a look.

After the first question is done, the second question does not need to be counted.

Because A2 = 2 and C = 1, the right alignment (denoted L) passes through the point P.

Let the perpendicular feet of a,b with respect to l be q,r, then by the definition of the conic curve, there is.

aq/br = af/af

By parallel lines, there is.

aq/br = ap/bp

Again by the symmetry of a, c has.

ap = cp,af = cf

bp/cp = bf/cf

bp/bf = sin∠bfp/sin∠bpfcp/cf = sin∠cfp/sin∠cpfsin∠bfp = sin∠cfp

But bfp = cfp can't be true, otherwise bfp is congruent with cfp.

BFP + CFP = 180 degrees, three-point collinear, certified.

First, the equation for simplifying the circle (x-1) +y-3) =11 (x-5) +y-6) =61-m

1): When the distance between the centers of the two circles is equal to the sum of the half-radii of the two circles, the two circles are inscribed to 5= 11+ (61-m).

Solution m(2): When the distance between the centers of the two circles is equal to the difference between the radii of the two circles, the two circles are inscribed with 5=|√11-√（61-m)|

Solve m, the tangent is perpendicular to the straight line where the center of the two circles is located, find the slope of the tangent, and after bringing m in, find the tangent point of the two circles, and the tangent can be calculated.

Bring in m, draw a picture, calculate the radius of the garden, the distance from the center of the circle, and the relationship with the tangent.

You can take a closer look at this problem and suspect that there is something wrong with the equation of the first circle, otherwise the calculation of the problem will be more complicated!

(1) Circle (x-1) 2+(y-3) 2=11 and circle (x-5) 2+(y-6) 2=61-m

inscribed, then, the distance from the center of the circle o1 to the center of the circle o2 is r1+r2

So |o1o2|=r1+r2

i.e. (5-1) 2+(6-3) 2=(11 =>m= 61-(5-11

2) When incising, |o1o2|=r2-r1 r1=11^<5=|o1o2|

5=-11^ =>m=61-(5+11^

Metric tangent (x-1) 2+(y-3) 2=11 - x-5) 2+(y-6) 2=61-m => 8x+6y=61-m-11+1+9-25-36=-m-1

3) Straight line 8x+6y=-m-1=-46 r2=(61-m) d=2*(4 2-(5 2) 2).

(1) AD is the height of the BC side, ADC=90

dac+∠c=90º

Similarly, CBE+ C=90

dac=∠cbe

again dac+ c=90

dac=90º-∠c

∠dac=∠cbe

AD, BE is the height of the side of BC and AC.

So dac+ c=90°

cbe+∠c=90°

So dac= cbe

In abc, abc=45°, bac=75°, so c=180°- abc- bac=180°-45°-75°=50°

Because dac+ c=90°

So dac=90°-50°=40°

Hello, Potato Group Shao Wenchao answers your questions.

If you don't understand anything about this question, you can ask it, and if you are satisfied, remember to adopt it.

It is not easy to answer the question, please understand, thank you.

And good luck with your learning!

As can be seen from the method of constructing the set of existential points you gave, use k1, k2, k3 to construct a basic trilateral, and then take the first point on any edge or its extension, and then make a straight line parallel to k1, k2, k3, and cross the basic trilateral or its extension line, and wait until a total of 6 points. These 6 points are the set of points that meet the requirements.

In fact, the basic trilateral is the set of points you require, but it is not in the ordinary sense that it meets the requirements. Each vertex of the trilateral has only 2 slopes, and if you think that there is a third slope in itself, each point can be constructed into a straight line with itself of any slope. From this point of view, the smallest set of points for n points is n.

If you put aside the construction of the straight line by itself, it will naturally take 2n points that are different from yourself. And these points are all present on the extension of the trilateral, 2 on each side. Altogether 2n.

And the opposite sides of the 6 sides are parallel.

For the case of 4 points, we can take a regular 8-sided shape as an example, and its slope of all diagonals and edges is greater than 4, which is satisfactory. If it is an octagon, if the opposite sides are parallel to each other, and each point naturally has 2 slopes, then the other 2 slopes must be obtained from its diagonal, then at least 2 diagonals must be parallel to the other 2 opposite sides. Such a structure does not seem to exist.

The solution of the sensory problem requires the concept of abstract geometry, such as the slope of any two points that meet the requirements, and think that there is a straight line between the two points, and vice versa, that the two points are not in a straight line. Or use the concept of groups in abstract algebra to start with the primitive form and expand it. The method of projective geometry at the same time may be more straightforward and helpful, but I wonder if the problem is related to Pappus's theorem, or if the infinity point may be considered more profoundly.

I don't know how to look at the basis and correlation in general linear space.

In general, the level of mathematics is limited, so it can be regarded as a brick throw.

I'm guessing what you mean by "the area of the shaded part" is: the sum of the areas of the four right triangles.

Let : ab = bc = cd = da = a, then: ae = bf = cg = dh = a 3; eb = fc = gd = ha = 2a/3.

The area of the square ABCD is equal to: A2

The area of the four right triangles (AEH, BFE, CGF, DHG) is equal to: 4*(a 3)*(2a 3) 2 = a 2*(4 9)

Answer: The ratio of the area of the shaded part in the figure to the area of the square ABCD is: B, 4/9.

Hehe, what kind of picture is this picture of the landlord, the geometric sketchpad didn't succeed.

Which is the shadow area?

Set the square side length = x x^2+(2x)^2=oa^2=5^2.Square ABCD area = 5

Could you please describe the problem clearly?

Connect O, A, the angle OAB is a right angle, set the square side length to X, it will be solved.

Look at the picture and speak.

The radius of the great circle is 8

The radius of the small circle is.

Then, in the last remaining iron sheet, one can not be intercepted with the same radius as the small circle, and the maximum radius can be intercepted.

The calculation method depends on the connecting line, and you can list the equation yourself.

The radius of the large circle is 8cm, and the radius of the small circle is 4cm. It is just enough to be able to intercept a circle that is the same as a small circle.

1. Draw a sketch of the known conditions.

2. Find the diagonal length of the square.

3. A perpendicular line that crosses the center to make a straight line.

4. According to the side length formula of a right triangle: half of the length of the square side and half of the length of the diagonal side find the length of the other side, that is, the length from the center to the side of the square.

5. According to the length obtained in 4, determine the point where one side of the square passes on the perpendicular line in 3, and combine the parallel straight lines to make the side of the orthogonal shape.

6. Make the other sides of the square.

7. I won't talk about finding the equations on each side, and determine a straight line according to the two points.

2x+y=0

2x+y+10=0

x-2y+5=0

x-2y-15=0

Look at my hard work, do you have a share.