Math Is Fun Forum

According to http://en.wikipedia.org/wiki/Median_%28geometry%29#Other_properties any triangle with sides a, b, c and median lengths m_a, m_b, m_c the following is true:

If our medians have lengths (2/3)a, (2/3)b, (4/5)c then

which can only be true if a = b = c = 0.

The short answer is "one step at a time".  This is the list of things we need to do for this problem:

1. Find a formula for the equation of the tangent line at a given point on the parabola.
2. Find the points on this tangent line that form the endpoints of the projection of the latus rectum.
3. Find a formula for the distance between these two points.
4. Find the minimum possible distance over the region we are told to work with.

Start by noting that the parabola is symmetrical so we only need to consider half of it, we'll go with the top half.  This allows us to simplify the equation from y²=4x to y=2√x.  Taking the derivative gives us y'=1/√x.  So for a given x-coordinate 'c', the slope of the tangent line at (c, 2√c) is 1/√c.  This makes the full equation of the tangent line y= (1/√c)x + b.  To solve for 'b' we plug in our point: 2√c = (1/√c)c + b, b = 2√c - c/√c, b = c/√c.  So our tangent line is y= (1/√c)x + √c.

For step 2 we need to find the endpoints of the projection of the latus rectum onto the tangent line.  The endpoints of the latus rectum itself are (1, 2) and (1, -2).  If you aren't sure why this is, the article on Wikipedia can explain.  To find the projection of these points we need the lines that pass through these points and are perpendicular with the tangent line, which means their slopes will be -√c.  For the point (1, 2) we have 2= (1)(-√c) + b, b=2 + √c, so we need to find the intersection between y= (1/√c)x + √c and y=-x√c + 2 + √c.  Set them equal to get (1/√c)x + √c = -x√c + 2 + √c, x(√c + 1/√c) = 2, x = 2/(√c + 1/√c) = 2/( [c + 1] / √c) = (2√c) / (c + 1).  Solving for y gives us y = (2 + c√c + √c) / (c + 1), which finally gives us ( [2√c]/[c + 1], [2 + c√c + √c]/[c + 1] ) as the first endpoint of the projection.  Doing the same for (1, -2) gives us ( [-2√c]/[c + 1], [-2 + c√c + √c]/[c + 1] ).

For step 3 we us the standard formula for the distance between two points, which is

This actually turns out pretty well for us:

To recap, what we have so far is the formula for the length of the projection of the latus rectum onto the tangent line at the point (c, 2√c).  Now we want to find the minimum possible value over the region c ∈ [0, 1].  Taking the derivative gives us

We can see that there are no critical points in the region we are working with, so we just need to evaluate the endpoints, or c=0 and c=1.  Solving for those gives us 4/√1 = 4, and 4/√2 = 2√2.  So our final answer is that the minimum length of the projection is 2√2 and occurs at the point (1, 2).  Since the parabola is symmetrical it also occurs at (1, -2).

There are four ways to get 4 dice whose product is 36: 6-6-1-1, 6-3-2-1, 4-3-3-1, and 3-3-2-2.  For each of these you need to count the number of ways that they can occur.  For example, there are six ways of getting the first one: 6-6-1-1, 6-1-6-1, 6-1-1-6, 1-6-6-1, 1-6-1-6-, 1-1-6-6.  There are 24 ways of getting the second set, 12 ways of getting the third set, and 6 ways of getting the fourth set, for a total of 6 + 24 + 12 + 6 = 48.  There are 6^4 = 1296 total outcomes for four dice, so the probability is 48/1296 = 1/27.

A number is divisible by 4 if its last two digits are divisible by 4.  In our case this means our last two digits can be 12, 16, 32, 36, 52, 56.

If our last two digits are 12 then the first digit can be 3, 4, 5, 6, and the second digit can be 0, 3, 4, 5, 6.  This gives us a total of 4*4 = 16 legal numbers that end in 12.  This process is the same for the other possible combinations, giving us a final answer of 16*6 = 96.

A second method is to start by counting the permutations of four digits out of the seven possible, which is 840.  1/7 of those begin with 0 and are illegal, so after removing those we have 720 legal four digit numbers.  From here we break these numbers down into three different types: those that begin with an odd number, those that begin with 4, and those that begin with 2 or 6.  You can see that half of the numbers are of the first type, 1/6 are of the second type, and 1/3 are of the third type.

For the first type of number, since the first digit is odd we know that 1/3 of the numbers have an odd number as the 10's digit.  Out of these remaining numbers 2/5 have 2 or 6 as the 1's digit.  Overall this gives us 720 * 1/2 * 1/3 * 2/5 = 48 legal numbers that begin with an odd digit.

For numbers that begin with 4, 1/2 have an odd digit in the 10's position, and 2/5 of the remaining numbers end in 2 or 6, giving us a total of 720 * 1/6 * 1/2 * 2/5 = 24 numbers that begin with 4.

For numbers that begin with 2 or 6, 1/2 have an odd digit in the 10's position and 1/5 of the remaining numbers end in 2 or 6, giving us 720 * 1/3 * 1/2 * 1/5 = 24 numbers the begin with 2 or 6.

This gives us a final answer of 48 + 24 + 24 = 96.

He said using digits 1-9, so it would be 9*3*9 = 243 numbers.  Also juan, when you say you want to find the sum of their digits, do you mean the total sum of the digits of all 243 of those numbers?

There are three types of squares on a chess board: a) those with an adjacent square on all 4 vertices, b) those with an adjacent square on 2 vertices, and c) those with an adjacent square on just 1 vertex.  There are 36 squares of type a), 24 of type b), and 4 of type c).  If a type a) square is chosen first there is a 4/63 chance that the second square will be corner adjacent to it, if a type b) square is first then there is a 2/63 chance that the second square will be corner adjacent, and a type c) square has a 1/63 chance.  This gives us a total probability of (36/64 * 4/63) + (24/64 * 2/63) + (4/64 * 1/63) = 49/1008 = 7/144.

Try to solve for y:

For y to be an integer we need sqrt( (21x^2 - 9) / 10) to be an integer.  We can first see that 21x^2 - 9 must be divisible by 10, so the last digit of 21x^2 must be 9.  This occurs whenever the last digit of x^2 is 9, which occurs when the last digit of x is 3 or 7.  So we can substitute x = 3 + 10h, x = 7 + 10h where h is any integer.  This gives us

Similarly for 7 we get 210h^2 + 294h + 102.  Since this must be a perfect square we should be able to factor them into the form (ah + b)^2, but you can see that this is impossible, so there are no solutions.

Do you mean integer solution?

There are an equal amount of odd and even numbers in the set S, and since a, b, c, and d do not need to be distinct we can simply treat all four as having a 50/50 chance of being either odd or even.  Notice that |ad - bc| is even if ad and bc are either both odd or both even, and it is odd otherwise.  Notice also that ad (and bc) is even if either a or d is even, and it is odd if both a and d are odd.  So the probability that ad is odd is 1/2 * 1/2 = 1/4, and this goes for bc too.  So the probability that ad and bc are both odd is 1/4 * 1/4 = 1/16, and the probability that both are even is 3/4 * 3/4 = 9/16, which gives us a 1/16 + 9/16 = 10/16 = 5/8 chance that |ad - bc| is even.

Assuming that all points are equally likely to be chosen, the probability of P being in any particular section of the sphere is equal to the ratio of the volume of that section to the volume of the entire sphere.  If we assume the sphere has radius 1, then we want to find the probability that P is within 1/2 of the center, so the probability U that P is closer to the center than the surface is