High school math geometry questions are rewarded for high scores, and high scores are urgently solve

First, the following lemma is demonstrated:

If a+b=a1+b1 proves: to make a triangle abc, a1b1c1 satisfies the angle a=a, the angle b=b, and the angle c=180-a-b; Angle a1 = a1, angle b1 = b1, angle c1 = 180-a1-b1 = 180-a-b = angle cFrom the sinusoidal theorem, there is bc ac=sina sinb=sinina1 sinb1=b1c1 a1c1

and c1=c, so the two triangles are similar. So there are a=a=a1=a1 and b=b1

In the original problem, let the quadrilateral be abcd, and the circumscribed circle o and ab, bc, cd, da are cut to e, f, g, h, eg to x, then ae=ah, eb=bf, fc=cg, gd=dhBy the sine theorem, there is.

sinaxh/sinaxe=[ah*(sinahx/ax)]/[ae*(sinaex/ax)]=sinahx/sinaex

Similarly, sinfxc singxc=sinxfc sinxgc

Join OE, OF, OG, OH, they are perpendicular to their respective edges. and oe = og = oh = of, from the isosceles triangle to get the angle ofe = angle oef, angle ohf = angle ofhBy sinusoidal properties, there is sinahx=cos(ahx+-90)=cosohf, and the same is true for sinaex=cosoeg, sinxfc=cosofh, sinxgc=cosoge

substitution, get sinaxh sinaxe=cosohf cosoeg=cosofh cosoge=sinxfc sinxgc=sinfxc singxc

and axh+axe=fxc+gxc, from the lemma knows axh=fxc, axe=gxc, by the equal of the vertex angles, knows a, c, x collinear, the same as b, x, d collinear. From this, it is clear that the original proposition is true.

Cylinder: the radius is r, the height is 2h, and the radius of the receiving ball r=(r square + h squared) square, for example, r=3, h=5

Then r=5 cone: cone top angle 2, bus 2L, external ball radius r=l cos

Cuboid: length 2a, width 2b, height 2c, radius of the outside ball r = (a square + b square + c squared) open.

Triangular pyramid: triangular pyramid ABCD, with the triangle ABC as the bottom surface and D as the vertex (ABCD four points are based on the coordinate table.

show). 1. Find the outer center point O of δabc (i.e., the intersection point of the perpendicular line on any two sides);

Second, the perpendicular line of the ABC of the surface O is crossed;

3. Set a point E on the perpendicular line, so that ED=EA (or ED=EB, or ED=EC) can list an equation and find the coordinates of the point E;

Outside ball radius r=ea=eb=ec=ed

Multiple (four or more) pyramids: The base polygon must be regular polygonal (quadrilaterals can be square or rectangular.)

shape), 1. Find the outer center of the bottom polygon o (i.e., the intersection of the perpendicular lines on any two sides), if found.

If there is no external center, then this polyangular pyramid has no external ball;

Second, the point o to do the perpendicular line of the bottom surface;

3. Set a point E on the perpendicular line so that ed=ea (or ed=eb, or ed=ec...).can be listed.

to find the coordinates of the point e;

Outside ball radius r=ea=eb=ec=ed

Prism: 1. Find the outer center of the two bottom surfaces of the prism;

2. Connect the outer center of the two bottom surfaces and find the midpoint m of the line;

3. The line connecting the point m with any vertex of the prism is the radius r of the outside ball.

Frontline issues.

1 Positional Relationship (Definition).

Intersect: There is one and only one common point.

Parallel: There are no common points in the same plane.

Heterogeneous: Different in any one plane, there is no common point.

2 Axioms and Inferences [to memorize].

3 Test Center -- Angles formed by straight lines on different planes Right angles Perpendicular lines (perpendicular intersection) Distance between straight lines on different planes.

Method: Point selection (Frequently selected: endpoint, midpoint).

Translation (linear planarization of space).

Also pay attention to summarizing the theorems introduced in the usual exercises, which can save time when making choices and filling in the blanks

Second, the surface problem.

1 Positional Relationship (Definition).

Lines are within polygons: there are an infinite number of common points.

Lines are off-polygon: Intersect: There is one and only one common point.

Parallel: There are no common points.

2 Lines and planes are parallel.

Definition, decision theorem, If a is not included in , b is included in , a b then a

The property theorem, if a, a is contained in =b then a b (lines and planes are parallel, lines are parallel).

3 The line surface is vertical.

Similar to parallel definitions, judgments, properties, point-surface distances, and so on

Diagonal projection The angle formed by the line surface.

Projection, etc., oblique segments, etc.

Diagonal segments, etc., projections, etc.

The perpendicular segment is the shortest.

Three-perpendicular theorem, inverse theorem.

Three-sided surface problem [similar to line surface problem, leave it to you to sort it out yourself].

When learning solid geometry, you can use some models (cubes, cuboids, space quadrilaterals, triangular pyramids, etc.) to help us memorize axioms, theorems. Especially when judging true or false propositions, you can find counterexamples in these models to help you judge. 】

First of all, the side of this round table is a fan ring (that is, a fan minus a concentric small fan leftover), the inner ring arc length is 2 * * 28 = 56, the outer ring arc length is 2 * * 63 = 126, and the side length is (50 2 + (63-28) 2) (1 2) = 5 149. Let the central angle of this fan ring be , and the radius of the inner ring is r, then *r=56, r+5 149)=126, and r=4 149, =14* 149 is obtained. If you want to use a piece of trapezoidal paper to cut this fan ring as small as possible, the upper bottom of this trapezoid must be the line connecting the two ends of the inner ring of the fan ring, the lower bottom must be tangent to the outer ring and the middle point of the outer ring, and the two waists and the two sides of the fan ring are collined.

Therefore, the upper bottom of the trapezoid=2*r*sin( 2), the height is 5 149*cos( 2), and the two bottom angles are ( - 2=(1 2-7 149)* Other calculations must be done with the help of calculators, and the idea is roughly like this.

I don't know what you mean behind it.

The first two sentences can be understood in this way, because you know the radius and height of the upper and lower bottom surfaces, you can find the height of the original cone, and the method is as follows.

To make the section of the original cone, it is a triangular ABC. (I look at the top one as A) and then make the line of the radius of the upper and bottom surface intersect with the height at O, and intersect with the section at D, so this triangle ABO is similar to the triangle where D is the perpendicular line.

The data generation can be found.

Because I can't draw on the computer, I can only type it out, but I guess it's still possible that I can't read it, sorry.

Upper bottom = the circumference of the upper part of the table body.

Bottom = the circumference of the circle of the lower part of the table.

Hypotenuse = 50*50+63*63=

(1) The height of the corresponding large cone is 90, the bus length of the upper small cone is 4 149, and the bus length of the large cone is 9 149. (2) into a fan, corresponding to the central angle = 14 149(3) The bow length of the small fan is the minimum upper bottom of the trapezoid, which is (8 149) sin (7 149).

Using trigonometric functions, the minimum value of the bottom = (18 149) tan (7 149) can be obtained

To find the symmetry point, the key is to grasp two points, the center point is on the straight line, perpendicular to the straight line (i.e. the slope is -1), and the slope of this straight line is 1 3

Let the symmetry point be q(m,n), then there is.

3+m)/2-3*(5+n)/2+2=0(n-5)/(m-3)=-3

Simultaneous solution yields q(5,1).

Let the point q be the symmetry point (m,n).

Then there is a midpoint on the line: (m+3) 2+(n+5) 2+2=0 perpendicular to the line: (5-n) (3-m)=-3

q(5,-1).

Let the p point and the ellipse intersect at the two points of the orange brigade, a(x1,y1)b(x2,y2), and the coordinates of the two points are brought into the elliptic equation.

x1^2/16+y1^2/4=1

x2^2/16+y2^2/4=1

Subtract the two formulas, (x1+x2)(x1-x2)+4(y1+y2)(y1-y2)=0

k=(y1-y2) (x1-x2)=-x1+x2) 4(y1+y2)=-2*2 (4*2*1)=-1

The point oblique equation of the ab equation is y-1=-(x-2) 2

x+2y-4=0

Bring y=-x 2+2 into the elliptic equation.

2x^2-8x=0

x=0 or 4ab|= or judgment (k 2+1)|x1-x2|=2√5

A simple way to reprimand you is to cross the point f1 and f2 to make a circle tangent to the straight line l, you can make a bunch of two, the smaller one and the tangent point of l is the point that is sought to cover p (the other is p2), take any point on the line, compare it with the angle f1pf2 or f1p2f2, use the equal arc to equal the circumferential angle, and then compare the angle f1pf2 and f1p2f2, you can prove. l intersects with x-axis n pn 2=f1n*f2n (cutting the year line) pf1 pf2=pn f2n (similarity ratio) pf1 pf2=f1n f2n=1+3 (1 2) 2 Except for elliptical junior high school students who do not understand, all other junior high school knowledge is used.

The elliptic focus is f1 and f2 respectively, and the point p is on the straight line l:x-(root number 3)y+8+2 (root number 3)=0, when f1pf2 takes the maximum value, let the point p and the ellipse intersect at two points a(x1,y1)b(x2,y2) The coordinates of the two points are brought into the elliptic equation x1 2 16+y1 2 4=1 x2 2 16+y2 2 4=1 Subtract the two formulas, (x1+x2)(x1-x2)+4(y1+y2)(y1-y2)=0 k=(y1-y2) ( x1-x2)=-x1+x2) 4(y1+y2)=-2*2 (4*2*1)=-1 2 ab The point oblique equation is y-1=-(x-2) 2 x+2y-4=0 Bring y=-x 2+2 into the elliptic equation 2x 2-8x=0 x=0 or 4 |ab|=√k^2+1)|x1-x2|=2√5

I think it should be the intersection of a straight line and two straight lines, first find the symmetry point of the point p about the straight line ab, it is very simple, and it comes out after thinking about it, it is p'(2,2), and the point of symmetry of the point p with respect to the y-axis is: p''(-2,0), so the intersection of the line formed by these two points with the y-axis, and the intersection with ab is the answer.

p is symmetrical p (4,2)f, p is symmetric with respect to the y axis, f(-2,0)p, e, h, f are in a straight line, p (4,2),f(-2,0) is substituted into the f p equation to get y=x 3+2 3

Substituting h(0,y),e(x,-x+4) to get h(0,2 3),e(,word limit, no graphical representation can be provided.