The math master is coming, the math master is coming

The first question is an Olympiad question of a certain year, but the landlord seems to have written the question wrong.

The original question is: A and B two ships set sail at the same time 50km apart, and keep the speed of the ship unchanged, if they sail against the water at the same time, A can catch up with B after 5 hours of sailing 100km, if the two ships sail along the water, what is the time for ship A to catch up with B and the voyage through it?

Solution: Suppose the velocity of the water flow is x, the velocity of A is a, and the velocity of B is b, then the equation of the column is 5(a-x)=5(b-x)+50 to get a-b=10

When two ships sail down the water.

Time t=50 (a-b)=5

Distance s=5(a+x)>100km

Second question: 2 to the power of 3m + 10n is equal to the power of 2 to the power of 3m multiplied by the power of 2 to the power of 10n.

32 to the nth power is equal to 2 to the 5n power.

So, 2 to the power of 3m + 10n is equal to the cube of 2 to the power of m multiplied by the square of 32 to the power n.

Since the m power of 2 is a and the nth power of 32 is b, therefore, 3m + 10n power of 2 is equal to the cube of a multiplied by the square of b.

1. The triangle BCP is all equal to the triangle DCP, so the angle PBC is equal to the angle PDC

Because PE is perpendicular to PB in a quadrilateral PBCE, the angular EPB = 90°

The angle ECB is also a right angle, so the angle PBC + angle PEC = 180°

Angle pec + angle ped = angle dec = 180°

So angular PBC = angular ped

So angular ped = angular pde

So PD=PE

2. Connect BE

CE=1 2,BC=1=>BE= 5 2,From the first question, we can see that PB=PD=PE, so the triangle Pbe is an isosceles right triangle, so.

pb=√10/4

Make PF perpendicular to BC

The triangle pbe is an isosceles right triangle, be=(root number 5) 2, so pe=(root number 10) 4, and the triangle pde is an isosceles triangle, so ef=1 4, so pf=3 4

So pf=fc

And because in the quadrilateral PFCG, PF is perpendicular to FC, PG is perpendicular to GC, and the angle GCF = 90°

So the quadrilateral PFCG is a square. So pc=3 (root number 2) 4

So ap=(root2)-pc=(root2) 4

3. Follow the above figure.

The area of the quadrilateral PBCE = the area of the triangle PBC + the area of the triangle PEC.

Because the triangle PBC is all equal to the triangle PDC, PB=PD, the angle PBC=Angle PDC, and the angle PGB=Angle PFD

So the triangle PBG is all equal to the triangle pdf. So pf=pg, so the quad pfcg is still a square.

So pg=pf=pc (root2)=(ac-ap) (root2)=x (root2).

So gc=x (root number 2), so bg=1-x (root number 2).

So ef=df=bg=1-x (root number 2).

So ec=1-df-ef=3-2 (root number 2)x

So in the right triangle pdf, df=

So the above area =

Just do the math.

In the end, there is a calculation, of course, there may be a mistake in the calculation, you can calculate it yourself, the way of thinking is this.

Ha?! I don't understand!! Is this junior high school? Or is it from high school? I haven't studied it!! =v=||Excuse me. Sahaha. Ha?! Ha?!

Junior high school! If you want to say what year it is, let's use the method!

（1）.The goal is to prove that the angle ped = the angle pde, then we get pe = pd. Extend the PE cross BC extension line at F, then the angle CEF = angle PBF

Angular pbc = angular pdc, so angular pdc = angular cef = angular ped, and finally pe=pd

2) Is AP 2 4 right?

Set n years to catch up with the United States has.

Get (solution n= log( 3

n about = so China can catch up with the United States in 15 years.

a525973239, hello:

Now it's like this, you should have studied physics, even if you haven't studied physics, you have also learned to chase the problem, just use the model of chasing the problem, there is a starting point, there is a speed, see when you can catch up, so the equation is as follows, 3000 (1+10%) t = 9000 (1 + 2%) t, where t is the number of years needed, (such as t = 1 means 1 year later) the two sides are simplified to get, (1+, after the two sides take ln, become t, the solution is ( =ln3, look up the table to find the t value.