Famous sayings of mathematicians How do mathematicians get there

Numbers rule the universe. Pythagoras, the queen of mathematics, the queen of science. Number theory, the queen of mathematics. C. Gauss, God created integers. All the rest of the numbers are man-made. L. clonek, God is an arithmetical. Jacoby, a mathematician who is not very popular in poetry

It takes 1 minute, 2 minutes, 5 minutes and 10 minutes for a, B, C and D to cross the bridge. Because it's dark, they have to cross the bridge with the help of a flashlight, but there is only one flashlight, and the bridge can only bear the weight of two people at the same time. Now we hope to cross the bridge in the shortest time. How to arrange the order of crossing the bridge? What's the shortest time?

B send a and then come back: 4 minutes
C send B and then come back: 10 minutes
Finally, Ding and C: 10 minutes
Minimum time: 24 minutes

Is the swing of swing and pendulum translation or rotation?

The characteristic of translation is: in the process of motion, the line segments between two points on the rigid body always keep parallel with the original, the motion trajectories of all points on the rigid body remain exactly the same, the line between two points of any rigid body keeps the same direction, and the displacement, velocity and acceleration of each point are the same, which can be treated as a particle

On the direction of zero vector
There are several sentences in the textbook
The direction of zero vector is arbitrary
The zero vector is parallel to any vector
There is a question in the exercise
If the nonzero vector a and B have the same or opposite directions, then the direction of a + B must be the same as that of one of a and B
To sum up, the zero vector is parallel to any vector, that is, parallel to a and B. but the direction of the zero vector is not the same as one of a and B. how to define the direction of the zero vector in this question?

The direction of the zero vector is arbitrary
But if we put it together with other vectors, we can say that the direction of the zero vector is parallel to it
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As shown in the figure, AE ⊥ AB, AF ⊥ AC, AF = AC are known, and the verification is as follows: (1) EC = BF (2) EC ⊥ BF

Mathematical answer group for you to answer, I hope to help you
A condition AE = AB should be added
prove:
(1) AE ⊥ AB, AF ⊥ AC, so: ∠ EAB = ∠ CAF = 90 °
Because AE = AB, AF = AC, ∠ EAC = ∠ EAB + ∠ BAC = ∠ CAF + ∠ BAC = ∠ BAF, △ EAC ≌ △ BAE,
So: EC = BF, ∠ AEC = ∠ ABF
(2) In △ AEB, ∠ AEC + ∠ CEB + ∠ Abe = 90 °
So: ∠ ABF + ∠ CEB + ∠ Abe = 90 °, that is: ∠ CEM + ∠ MBE = 90 °,
Therefore, ∠ BME = 180 ° - (∠ CEM + ∠ MBE) = 90 ° in △ EBM
So: EC ⊥ BF
I wish you progress in your studies and make progress! (*^__ ^*)

1. In the triangle ABC, if (Sina) &# 178; + (SINB) &# 178; = sin (a + b), AB is an acute angle, find the value of a + B
(solved by cosine, sine and tangent formula of sum and difference of two angles)
2. If the two roots X1 and X2 of the equation x & # 178; + X · cos α cos β + cos γ - 1 = 0 satisfy X1 + x2 = X1 * X2, then the shape of the triangle with α, β and γ as internal angles is

1sin ^ 2 (a + b) = Sina ^ 2 + SINB ^ 2, sin ^ 2C = sin ^ 2A + sin ^ 2BA / Sina = B / SINB = C / sinc = ksina = A / K, SINB = B / K, sinc = C / KC ^ 2 / K ^ 2 = a ^ 2 / K ^ 2 + B ^ 2 / K ^ 2C ^ 2 = a ^ 2 + B ^ 2, triangle ABC is right triangle, Sina ^ 2 + SINB ^ 2 = 1 sin (a + b) = Sina ^ 2 + SINB ^ 2 = 1

As shown in the figure, if AB is parallel to CD, BF bisecting angle Abe, DF bisecting angle CDE, and angle bed = 75 degrees, then angle BFD =?
Swap the positions of E and F in the graph, hoping to get the process quickly,

Make parallel line EP of AB through point e according to the same internal stagger angle, so angle Abe = angle BEP, angle CDE = angle DEP, because BF bisects angle Abe, DF bisects angle CDE, so angle FDE + angle FBE = angle Abe / 2 + angle CDE / 2 = angle BEP / 2 + angle dep / 2 = 75 / 2 = 37.5 degrees, connect Fe into two triangles, so angle BFE + angle EFD + angle Feb + angle fed + angle EBF +

Suppose there are three points a (- 2,2) B (- 1,4) C (4, - 5), and the vector AB = (1 / 2) vector CD, then the coordinates of point D are obtained

Let D (x, y)
Vector AB = (1,2) vector CD = (x-4, y + 5)
Because the vector AB = (1 / 2) the vector CD
So 1 = 1 / 2 (x-4) 2 = 1 / 2 (y + 5)
x=6 y=-1

Exploration: in Figures 1 to 3, the area of △ ABC is known as a. (1) as shown in Figure 1, extend the edge BC of △ ABC to point D, so that CD = BC, connecting da. If the area of △ ACD is S1, then S1= (expressed by the algebraic formula containing a) (2) as shown in Figure 2, extend the edge BC of △ ABC to point D, extend the edge CA to point E, so that CD = BC, AE = Ca, connect de. if the area of △ Dec is S2, then S2= (expressed by the algebraic formula containing a) (3) on the basis of Figure 2, extend AB to point F, make BF = AB, connect FD and Fe, and get △ def (as shown in Figure 3). If the area of the shadow part is S3, then S3= (expressed by the algebraic formula containing a), and use the conclusion of (2) to write the reason. It is found that: as above, the edges of △ ABC are extended by one time in turn, and the end points are connected to get △ def (as shown in Figure 3). At this time, we call △ ABC expanded outward once. We can find that the area of △ def obtained after one expansion is the same as that of the original △ ABC Application: to plant flowers in a large enough open space, the engineers designed the following patterns: first, plant red flowers in the open space of △ ABC, and then expand △ ABC outward three times (the patterns of the first two expansion are shown in Figure 4). Plant yellow flowers in the first expansion area, purple flowers in the second expansion area, and blue flowers in the third expansion area The area of △ ABC is 10 square meters. Please use the above conclusion to calculate: (1) the area of purple flower; (2) the area of blue flower

(1) The height of BC and CD is equal, BC = CD, according to the equal area of the triangle with equal base and equal height, S1 = s △ ACD = a, so the answer is: A. (2) connect ad, similar to (1), according to the equal area of the triangle with equal base and equal height, s △ ACD = s △ ade = a, ∧ S2 = 2A, so the answer is: 2A. (3) similar to (2): s △ AFE = s △ BFD = s △ CDE = 2A, ∧ S3 = 2A + 2A+ 2A = 6a, so the answer is: 6A. (3) the area of yellow flower area is 6 × 10 = 60 square meters, the area of purple flower area is 6 × (60 + 10) = 420 square meters; the area of blue flower area is 6 × (420 + 60 + 10) = 2940 square meters

F1F2 is the left and right focus of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), and a is the vertex of ellipse C
B is a point of intersection between straight line af2 and ellipse C, ∠ f1af2 = 60 °
(1) Find the eccentricity of ellipse C
(2) Given that the area of triangle af1b is 40 √ 3 (40 root sign 3), find the value of a and B
I figured out that the eccentricity was 1 / 2
The second question thinks it can be solved by polar coordinate method, but it can't be solved. Notice that I'm seeking polar coordinate method

The one upstairs is incomprehensible. To answer the second question, I don't know how to pretend to understand
(1) The first is the definition of an ellipse that shows 2A = 4, α = 2, Ca (1,3 / 2) on the ellipse that can be obtained by B 2, and the point that a = 2 brings into the elliptic equation = as shown in Figure 3, the elliptic equation x 2 / 4 + y 2 / 3 = 1, C = √ 2-B 2 = 1, then the coordinates of focus F1, F2 (1,0), (- 1,0) (2) let the coordinates of M (x1, Y1), the coordinates of P (X2, Y2), m, n be symmetric to the origin, and the coordinates of n (- x1, - Y1) =KPN of (y2-y1) / (x2-x1) = (Y2 + Y1) / (x2 + x1), then KPN of KPM * = (Y2 ^ 2-y1 ^ 2) / (× 2 ^ 2 - X1 ^ 2)
P point m is on an ellipse, and there are two points
X1 ^ 2/2 + Y1 ^ 2 / B 2 = 1①
2 times ^ 2 / 2 + Y2 ^ 2 / B 2 = 1 ②
② - one
(× 2 ^ 2-x 1 ^ 2) / 2 + (Y2 ^ 2-y 1 ^ 2) / B 2 = 0, that is, KPN of KPM * = (Y2 ^ 2-y 1 ^ 2) / (× 2 ^ 2-x 1 ^ 2) = - B 2 / 2, so KPN of P KPM * is position independent
The hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) has similar characteristics
KPN = B 2 / a 2 of KPM * proves that in the case similar to the previous case of ellipse, the sign is changed back, i.e
X1 ^ 2/2 Y1 ^ 2 / B 2 = 1①
2 times ^ 2 / 1-2-y2 ^ 2 / B2 = 1 ②
② - 1
（×2 ^ 2-X 1 ^ 2）/ 2 - （Y2 ^ 2-Y 1 ^ 2）/ B 2 = 0
That is to say KPM * KPN = (Y2 ^ 2-y1 ^ 2) / (x2 ^ 2-x1 ^ 2) = B2 / A2, so KPM * KPN is independent of position

Famous sayings of mathematicians How do mathematicians get there