Difficult math and geometry problems in junior high school.

I am a junior high school student in a small county in Jiangsu Province, and I have always been good at mathematics in primary school, but I have slowly slipped since the first year of junior high school, I don't listen carefully in class, I can't do many math problems, and my grades are hovering at the passing line, and I slowly develop into an aversion to mathematics, and I 、... in the first year of junior high school

There is something wrong with this question.

From the diagram, point D is outside of ABC. But the question says that d is the point in abc. There are contradictions.

As shown in the figure, the triangle plate ABO containing a 30° angle is placed into the plane Cartesian coordinate system, and the coordinates of a and b are (3,0) and (0,3) respectivelyThe moving point p starts from point A and moves along the polyline ao-ob-ba, and the velocity of the point p on ao, ob, ba is 1, ,2 (length unit seconds) and the upper edge of the ruler l moves upward parallel at a speed of 33 (length unit seconds) from the position of the x-axis (i.e., the l x-axis is maintained during the movement), and intersects with ob and ab at two points e and f respectively, and the moving point p and the moving line l start at the same time, and the motion time is t seconds, when the point p moves along the polyline ao-ob-ba for one week, The linear line l and the moving point p stop moving at the same time

Please answer the following questions:

1) The analytic formula for a straight line passing through a and b is

2) When t 4, the coordinates of the point p are when t point p coincides with the point e;

3) Make the point p with respect to the symmetry point p of the straight line efIn the course of motion, if the formed quadrilateral pep f is a diamond, what is the value of t?

When t 2, there is a point q such that feq bep ?If it exists, find the coordinates of the point q;

If not, please explain why

This question is the last question in the middle of the test, and it is difficult

See **,m and n in **, which are l and k in this question

E DO ED BC, DK BC, EF PH DK IN TRAPEZOIDAL EFKD.

p is the midpoint, so h is the mn midpoint.

ph=1/2（ef+dk）

When doing eg ac, then eg = ef

In EGD, PN is the median line.

pn=1/2eg

Do do ab over d

In the ODE, it can be proved that PM is the median od=dk

pm=1/2od

pm+pn =1/2od+1/2eg=1/2（ef+dk）ph=pm+pn

The picture is more annoying.,I did it directly with the drawing board.,Take a screenshot for you to see.。

ce=df+cg

After D is the perpendicular line of DE, the extension line of EF is crossed with H, which proves that EDH is equal to ABG, and then proves that DF=DH, so that DF=BG=BG=CE

Over e for EM vertical AD, m for the vertical foot.

de=2em

em＝x2x）²－x²＝（a／2）²

3x²＝a²／4

Find: It is easy to get ae=de=ef

So what is needed is:

Extend DE to AB and G, Extend EF to AD to H, and BC to I.

Angle AEG = EAD + EDA = 30 + 30 = 60 [Outer Angle Theorem], Angle EAG=90-EAD=60

Angle ega = 180-aeg-eag = 60 [eag inner angle and 180].

EAG is an equilateral triangle [60 at all three corners].

ag=ae=eg=ed [The triangle EAD is an isosceles triangle, AE=ED].

In the right-angled triangle GAD, let ag=b, dg=2b, and the Pythagorean theorem obtains, AD=root number3B=A, B=A, root number3

ae+ed=2a root number 3, the same way bf+cf = 2a root number 3

eh=fi [easy to prove congruence, omitted] e is the midpoint of eg, eh is the median of the triangle dag, eh=1 2ag [median line theorem].

So eh+fi=2eh=ag

ef=ab-fi-eh=ab-ag=a-b

Total wire length = a-b + 2a root number 3 + 2a root number 3 = a + 3a root number 3 = a + (root number 3) a

Note: Add and supplement by yourself].

Let de=x, ad=root number 3x (pass e to make a perpendicular line, 30 angles rt triangle, this you must have solved in junior high school).

Therefore x = root number 3a 3

Next, all kinds of perpendicular lines can be added to get ef=a(1-root number 3 3).

Feel free to ask.

E is EM vertical AD, m is perpendicular foot, F is fn vertical BC, n is perpendicular foot and EM x ade= aed=30

de=2em

It is obtained according to the Pythagorean theorem.

2x)²-x²=(a/2)²

3x²=a²/4

x=√3a/6

ef=(1-√3/3)a

The required wire length is (5-3 3)a

Over e for EM vertical AD, m for the vertical foot.

de=2em

em＝x2x）²－x²＝（a／2）²

3x a 4 x a root number 3 68x a 2 4a root number 3 3 a 2

The triangle ADE is an isosceles triangle, the AD length is A, the midline length is 3A 6, and the AE is 3A 3, and the same way for the triangle BCF, the EF length is (1- 3 3)A, so EF+4*AE is required

ae=ed=bf=cf=a/2

EP is the perpendicular line of AD, and FO is the perpendicular line of BC.

ep=fo=√(ae)^2-(ap)^2

ef=a-ep-fo=2√3

The question seems to be wrong, or it is not explained.

I also think the inscription is wrong.