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2)   Let's place a point P on a circle opposite to point M. Now we have a right triangle CMP

∠ABC = arctan( AC / BC )

∠CMP = ∠ABC - ∠BCM

CM = cos(∠CMP) * MP

What is the circumradius of a triangle ABC with sides of length 6,8,and 10?  

We recognize that a 6, 8, 10 triangle is a right triangle.  6² + 8² = 10²  

Since it's a right triangle, the hypotenuse is the diameter of the circle it's inscribed in. 

The radius is half the diameter, so ....  radius = 5  

_.

Right scalene Pythagorean triangle.
Sides: a = 6 b = 8 c = 10

Area: T = 24
Perimeter: p = 24
Semiperimeter: s = 12
Circumradius(R)
The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect.

R=(abc) / 4rs =6.8.10 / 4.2.12 =5 

The circumcenter of a right triangle occurs at the midpoint of the hypotenuse.

The radius of the circumcircle will be the distance from the midpoint to either end of the hypotenuse;

in this case, the radius is 5.

If you find the vertex, you will have the highest point.

One way to find the vertex is to complete the square:

         y  =  -12x² + 48x

         y  =  -12(x² - 4x)         

         y  =  -12(x² - 4x + 4) + 48

         y  =  -12(x - 2)² + 48

When  x = 2  the height will be  y = 48.

3ax² + 3bx + 3c

Factor out the 3:           3[ ax² + bx + c ]

Factor out the a:           3[ a( x² + (b/a)x ) + c ]

Complete the square:  3[ a( x² + (b/a)x + b²/(4a)² ) + c - b²/(4a) ]

Factor:                         3[ a( (x + b/(2a) )² ) + ( c - b²/(4a) )]

Multiply back the 3:         3a( (x + b/(2a) )² ) + ( 3c - 3b²/(4a) )

n = 3a         h  =  -b/(2a)     k  =  3c - 3b²/(4a)

First we find the number of terms which is (103-1)/3+1=102/3+1=34+1=35. Now notice that every pair adds up to 103+1=104. So the sum is 104(35)/2=1820.

1 - How many terms divisible by 2 are there between 2 and 8??

You subtract the last from the first: 8 - 2 =6. This is the number of terms.

6 / 2 = 3. But 3 doesn't count both 8 and  2. When you are subtracting from each other, only one of them is

counted, so you must add 1 to count both of 8  and 2. So that you have: 2, 4, 6, 8 for a total of 4.

Exactly the same thing with 1,000,000,001 as the first term and 9,9999,999,999, the last term.

9,999,999,999 - 1,000,000,001 =8,999,999,998 /2 =4,499,999,999 + 1 =4,500,000,000.

Got it?

Thank you so much! I looked at the original post but it was a tinypic photo so it didn't help me at all. Now time to go and finally understand this problem... Thanks again! 

:)

@answer #1 -- Thanks for your answer! However, I'm not sure I understand what you did! 

I get that the smallest and the largest numbers that meet are the ones you listed. However, why are you subtracting them, then dividing that by 2 and adding 1? I'm confused, can you explain? 

FYI for future answerers, I figured out #3, and #4, I don't need help and those. I still need help on #1 and #2. Thanks! 

I think what s/he meant was

If m is an integer is \(m^3\cdot 3^{k+3}-m^2\cdot 3^{2k+3}+m\cdot 3^{k+2}-1\) an integer?

Btw, try to do the writing problems by yourself on AoPS. They are a refelection of how well you learned and you should try to avoid getting help from other websites, as you can always post on the message board.

-C

Sorry for the typo!. The total should be = 12

y  =  -6t² + 51t

y = 108

Combining:                        -6t² + 51t  =  108

                                 -6t² + 51t - 108  =  0

Multiplying by -1:      6t² -  51t + 108  =  0

Dividing by 3:              2t² - 17t + 36  =  0

Factoring:                      (2t - 9)(t - 4)  =  0

Answer:  Either  t = 4.5  or  t = 4

(1,  4,  9,  16,  36,  64,  81, 144,  324, 576, 1296, 5184) >>Total = 13

For the first one:

1)  Extend QRS to point T on the circle.

2)  By the "power of a point":  QP²  =  QR·QT   --->   6²  =  3·QT   --->   QT = 12

      Since QR = 3, RT = 9.

      Since RS = 3, ST = 6.

3)  Draw OT and OR, creating triangle(TSO) and triangle(RSO).

      OT and OR are radii; Let the radius = r.

4)  Use the Law of Cosines on these two triangles.

      OT²  =  r²  =  6² + 2² - 2·6·2·cos(TSO)

      OR²  =  r²  =  3² + 2² - 2·3·2·cos(RSO)

      Since these are equal:  6² + 2² - 2·6·2·cos(TSO)  =  3² + 2² - 2·3·2·cos(RSO)

                                                        40 - 24·cos(TSO)  =  13 - 12·cos(RSO)

      Since angle(RSO) is supplementary to angle(TSO),  cos(RSO) = - cos(TSO)

           --->                                       40 - 24·cos(TSO)  =  13 + 12·cos(TSO)

                                                                                27  =  36·cos(TSO)

                                                                              0.75  =  cos(TSO)

5)  Since  OT²  =  r²  =  6² + 2² - 2·6·2·cos(TSO)

           --->            r²  =  40 - 24·0.75  

                            r²  =  22

                             r  =  sqrt(22)

Thanks!

Hello fellow AoPSer!

For the first one you should extend QS such that it intersects the circle again at T. And extend OS to intersect the circle at U and V. Try to find ST from there and set variables to solve an equation with power of a point.

For the second one. Let the intersection of AB and MC be D. Drop an altitude from D to AC. Then use the Angle Bisector Theorem, Similar Triangles, and lastly Power of a Point to get your answer.

Hope this helped!

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Can be done as follows: