Tiggsy

Angles A and B are each 90 degrees, and given that the sides of the pentagon are of equal length, (draw a diagram,) it follows that ABCE form a square. The other two sides of the pentagon CD and DE, together form an equilateral triangle with CE.

The angle E is made up of 90 degrees from the square and 60 degrees from the equilateral triangle, 150 degrees in all.

Suppose that on the trip from the cafe to the boba shop a total of U metres is uphill, ( either in a single stretch or made up in bits), and similarly L metres is on the level, and D metres is downhill, then the time taken for the trip will be U/100 + L/120 + D/150.

On the return trip uphill becomes downhill and downhill becomes uphill so the time taken will be U/150 + L/120 + D/100.

Total time there and back will be U(1/100 + 1/150) + L(1/120 +1/120) + D(1/150 + 1/100) = (U + L + D)/60.

The total distance covered is 2(U + L + D) metres, so dividing distance by time, the average speed will be 120 m/hr.

6  3  9

1  8  7

1  1  2

No other possibilities.

Looks to me that E is closer to the side of the triangle than it is to the point A.

(Same goes for D and F.)

Much easier if you notice that this is a quadratic X^2 + X - 12= 0, where X = (4x -16)/(3x - 4).

Solve for X (= -4 or 3 ) and then for x.

Two possible methods.

1. Calculate the lengths of AB, BC, CA and use the cosine rule to calculate angle CBA, then use the formula

area = BA.BC.sin(CBA)/2 to calculate the area. That requires you to calculate the lengths of the three sides..

2. Let the angle that CB makes with the x-axis be alpha, the angle that AB makes with the x-axis beta, then angle CBA, call it gamma, will be beta - alpha. From that

\(\displaystyle \tan\gamma=\tan(\beta-\alpha)=\frac{\tan\beta-\tan\alpha}{1+\tan\beta.\tan\alpha}. \)

\(\tan\beta \text{ and }\tan\alpha\)

are the slopes of the lines AB and CB respectively.

The area of the triangle can be calculated as the area of the trapezium minus the areas of two triangles.

If you are really keen, do it both ways and see that the answers are the same.

Correct up to there, just two more steps.

Let, 

\(\displaystyle a = m^{2}-n^{2},\\\displaystyle b=2mn,\\ \displaystyle c=m^{2}+n^{2},\)

then a, b and c  form a Pythagorean triple.

\(\displaystyle a^{2}+b^{2}=(m^{2}-n^{2})^{2}+(2mn)^{2} \\ \displaystyle =m^{4}+n^{4}-2m^{2}n^{2}+4m^{2}n^{2} \\ \displaystyle =m^{4}+n^{4}+2m^{2}n^{2}=(m^{2}+n^{2})^{2} \\ \displaystyle = c^{2}.\)

What we are looking for then are values of m and n such that m^2 + n^2 = 97.

m^2  = 97 - n^2, so try

n =1 , m^2 = 97 - 1 = 96, m not an integer,

n = 2, m^2 = 97 - 4 = 93  mnai,

n = 3, m^2 = 97 - 9 = 88  mnai,

n = 4, m^2 = 97 - 16 = 81, m = 9, got it.

So, a = 9^2 - 4^2 = 81 - 16 = 65,

b = 2*9*4 = 72.

Check, (for silly mistakes), a^2 + b^2 = 65^2 + 72 ^2 = 4225 + 5184 = 9409 = 97^2. 

Run a horizontal through P to meet the circle again at A.

Drop a vertical through P to meet the circle again at B.

Angle APB is a rt-angle so AB will be a diameter of the circle.

To calculate the length of PA, drop a vertical from the point of contact with the wall onto PA and show that half the length of PA is           r - 4, so  that the length of PA is 2r - 8, where r is the radius of the circle.

Similarly the length of PB is 2r - 10. 

Now, by Pythagoras in APB, (2r - 8)^2 + (2r - 10)^2 = (2r)^2,

leading to

r^2 - 18r +41 = 0,

and r = 15.325 (3 dp).

Two methods.

 Using CPhill's diagram,

E, H and D are collinear, (the line EHD is at rt-angles to the common tangent at E).

Let the radius of the small circle be c, then from the triangle GHD, c^2 + (1/2)^2 = HD^2,

so HD = sqrt(c^2 + 1/4),

so ED = HD + c = sqrt(c^2 + 1/4) + c = 1, (the radius of the big circle).

c^2 + 1/4 = (1 - c)^2 = 1 -2c + c^2,

2c = 3/4,

c = 3/8.

Let the radius of the small circle be c, then its Cartesian equation is

x^2 + (y - c)^2 = c^2

x^2 + y^2 -2yc = 0.................(1)

The equation of the big circle centre D is

(x - 1/2)^2 + y^2 = 1

x^2 - x + y^2 -3/4  = 0..................(2)

Solving simultaneously, (subtract (1) - (2)),

x - 2yc + 3/4 = 0.

Substitute for x in (1),

(2yc - 3/4)^2 + y^2 - 2yc = 0,

4y^2c^2 - 3yc + 9/16 + y^2 - 2yc = 0,

y^2(4c^2 + 1) - 5yc + 9/16 = 0.

For tangency, the equation needs to have a single root, so

25c^2 - 4(4c^2 +1)(9/16) = 0,

16c^2 - 9/4 = 0,

c^2 = 9/64,

c = 3/8.

\(\displaystyle 1 + i\sqrt{3} = 2(1/2 + i\sqrt{3}/2)=2(\cos(\pi/3)+i\sin(\pi/3)). \\ (1+i\sqrt{3})^{6}=\{2(\cos(\pi/3)+i\sin(\pi/3))\}^{6}=2^{6}(\cos(2\pi)+i\sin(2\pi)) \\ =2^{6}=64.\)

sin(theta) = cos(5) - sin(25) = sin(85) - sin(25).

Now use the trig identity for  sin(A) - sin(B).

The line of symmetry passes through the minimum point on the curve and that will be at x = -b/2a, so -b/2a = 2, 

4a + b = 0 .....................(eq 1).

The curve passes through (1, 1), so

a + b + c = 1 .........................(eq 2)

and through (4 - 7) so

16a + 4b + c = -7 .....................(eq 3).

Solve the three equations simultaneously, begin by subtracting 4 times eq 1 from eq 3, to arrive at 

a = -8/3, b = 32/3, c = -7.

Putting y = 0 and solving the resulting quadratic,

\(\displaystyle x = \frac{-32/3 \pm \sqrt{(32/3)^{2}-4(-8/3)(-7)}}{2(-8/3)} \\ \displaystyle =2 \mp(3/16)\sqrt{352/9} = 2 \mp(3/16)(4/3)\sqrt{22} \\ \displaystyle 2 \mp\sqrt{22/16}= 2 \mp \sqrt{11/8},\)

so n = 11/8.

cosxcosy - sinxsiny = cos(x + y) = 0.6 - 0.2 = 0.4,    x + y = arccos(0.4).

cosxcosy + sinxsiny = cos(x - y) = 0.6 + 0.2 = 0.8,   x - y = arccos(0.8).

etc.

Answered a few days ago, here.

web2.0calc.com/questions/arithmetic-geometric-series#r3