Mathematician's story (to be brief)

Mathematician's story (to be brief)

The story of a scientist Hua Luogeng
Hua Luogeng was born in Jintan County, Jiangsu Province on November 12, 1910. His family was poor and he was determined to study hard. When he was in middle school, in a math class, the teacher gave the students a famous problem: "there is a number, three, three, two; five, five, three; seven, seven, two. How much is this number?" when we were thinking about it, Hua Luogeng stood up and said: "23" his answer surprised the teacher and was praised by the teacher. From then on, he fell in love with mathematics
After finishing the first grade of junior high school, Hua Luogeng was out of school because of his poor family, so he had to stand at the counter for his parents, but he still insisted on self-learning mathematics. Through his unremitting efforts, his paper "the reason why the solution of the quintic equation of Su Jiaju's algebra can't be established" was discovered by Professor Xiong Qinglai, director of the Department of mathematics of Tsinghua University, and invited him to Tsinghua University. Hua Luogeng was employed as a university teacher, This is the first time in the history of Tsinghua University
In the summer of 1936, Hua Luogeng, an outstanding mathematician, worked in Cambridge University for two years as a visiting scholar. At this time, the news of Anti Japanese war spread all over Britain. With strong patriotic enthusiasm, he went back to his motherland and gave lectures for Southwest United University
Hua Luogeng paid great attention to the direct application of mathematical methods in industrial and agricultural production. He often went to factories for guidance, popularized the application of mathematics, and compiled popular science books

How to divide a 30 degree right triangle into four identical triangles
Attention! It's four!

Take the middle point of the hypotenuse of a right triangle and make a vertical line to the two right sides
Then connect the intersection of the middle point and the two right angle sides, and the four triangles are congruent

Elevator lifting motion belongs to translation, rotation

Translation

It is known that if the vector AB = A-B, the vector AC = 2a-b, the module of a = 3, the module of B = 4, and the angle between a and B is 60 °, the lengths of the three sides of the triangle ABC are

AB=√13 BC=3 AC=2√7

As shown in the figure, in △ ABC, ab ＞ AC, e is the midpoint of BC side, ad is the bisector of ∠ BAC, through e is the parallel line of AD, intersecting AB at F, intersecting the extension line of Ca at G

It is proved that: extend Fe to Q so that EQ = EF, connect CQ, ∵ e is the midpoint of BC side, ∵ be = CE, ∵ be = CE, ∵ be = ceqef = EQ, ≌ CEQ, ≌ BF = CQ, ∵ BFE = q, ∵ ad bisecting ∵ BAC, ∵ CAD = bad, ∵ EF ≌ ad, ≌ CAD = g, ≌ bad =...., be = CE, be = be = CE, be = beqef = EQ, ≌ BF = CQ

In the cube abcd-a1b1c1d1, where e is the midpoint of dd1, what is the angle between BD1 and ace?

Connect BD, AC to o, OE
In △ dd1b, O and E are the midpoint of BD and dd1 respectively
That is, OE / / = 1 / 2bd1,
In addition, OE is included in plane EAC (if it is consistent, it is not printed), and BD1 is not included in plane EAC
Planar eac
{BD1} plane ace
The angle between BD1 and ACE is 0 degree

Given that the triangle ABC is congruent with the triangle EFG, and ab = EF, BC = FG, angle a = 68 degrees, angle f-angle g = 56 degrees, find the degree of angle B and angle C
The process should be detailed!

Triangle ABC and triangle EFG congruence
So, angle B = angle F, angle c = angle G
So angle B-angle C = angle e-angle f = 56
From angle B + angle c = 180 - angle a = 112
therefore
Angle B = 84, angle c = 28

Given that vector E is unit vector, vector a multiplies vector E = - 2, and the angle between vector a and vector E is two-thirds π, then the projection of vector a on vector E is

The projection of vector a on vector E is vector A &; vector E / | vector E | = vector A &; vector E = - 2
(other conditions don't work)

As shown in the figure, in the triangle ABC, points D and E are the middle points of edge AB and AC respectively. It is proved that the area of triangle ade is 1 / 4 triangle
Did not learn the median line, similar

Obviously, the height of triangle ade from point a is half of the height of triangle ABC from point a, and the length of side De is half of that of side ab. from the fact that the open area of triangle is equal to half of the bottom multiplied by height, the area of triangle ade is 1 / 4 of that of triangle ABC

The distance between the point P on the ellipse x2100 + y236 = 1 and its left quasilinear is 10, then the distance between the point P and its right focus is 10______ ．

According to the meaning of the question: a = 10, B = 6, C = 8, e = CA = 45, so there is a right quasilinear equation: x = A2, C = 252, we can see from the definition of ellipse that the distance from point P to the left focus is 10 × 45 = 8, and the distance from point P to the right focus is 2a-8 = 12, so the answer is 12