Wednesday, October 10, 2012

Cool math Olympics - II

Cool math Olympics - 1960

1. Determine all three-digit numbers `N` having the property that `N` is divisible by `11`. and `N / 11` is equal to the sum of the squares of the digits of `N`.
2. For what values of the variable `x` does the following inequality hold: `(4x^{2})/(1 -\sqrt{1 + 2x})^{2} < 2x + 9`.
3. In a given right triangle `ABC` the hypotenuse `BC` of length `a` is divided into `n` equal parts (`n` an odd integer). Let `\alpha` be the acute angle subtending, from `A` that segment which contains the midpoint of the hypotenuse. Let `h` be the length of the altitude to the hypotenuse of the triangle.
         Prove: `tan(\alpha) = (4nh) / ((n^{2} - 1) * a)`.
4. Construct triangle `ABC` given `h_{a}, h_{b}` (the altitudes from `A` and `B`) and `m_{a}`, the median from vertex `A`.
5. Consider the cube `ABCDA^{1}B^{1}C^{1}D^{1}` (with face `ABCD` directly above face `A^{1}B^{1}C^{1}D^{1}`).
        (a) Find the locus of the midpoints of segments `XY` where `X` is any point of `AC` and `Y` is any point of `B^{1}D^{1}`.
        (b) Find the locus of points `Z` which lie on the segments `XY` of part (a) with `ZY = 2XZ`.
6. Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let `V1` be the volume of the cone and `V2` the volume of the cylinder.
        (a) Prove that `V1 != V2`.
        (b) Find the smallest number `k` for which `V1 = kV2`, for this case, construct the angle subtended by a diameter of the base of the cone at the vertex of the cone.
7. An isosceles trapezoid with bases `a` and `c` and altitude `h` is given.
        (a) On the axis of symmetry of this trapezoid, find all points `P` such that both legs of the trapezoid subtend right angles at `P`:
        (b) Calculate the distance of `P` from either base.
        (c) Determine under what conditions such points `P` actually exist. (Discuss various cases that might arise.)

Cool Math Olympics - 1959 Key

1. Solution: `GCD` of `a` and `b` is the greatest common divisor of `a` and `b`. It is known that, what divides `a` and `b` also divides `a - b`. So, `GCD` of `a` and `b` should be same as that of `a` and `a - b` (when `a > b`) or that of `b` and `a - b`. Considering that, and also from the observation that, some rational number is irreducible, iff, the numerator and denominator have a `GCD` of `1`, it follows that:

`GCD(21n + 4, 14n + 3) = 1`                                    <=>
`GCD(21n + 4, (21n + 4) - (14n + 3)) = 1`               <=>
`GCD(21n + 4, 7n + 1) = 1`

Now, calculate GCD normally, like we do, divide take remainder and repeat the procedure with the remainder and the divider. The remainder in `21n + 4` when divided by `7n + 1` is `1`. which means the `GCD` is `1`. It immediately follows that, the GCD of original numbers is also `1`, hence, those are irreducible.

2. Square the expression on both sides, noting that `A >= 0` (since square root is a function, and results of that function are always non-negative).

                  `\sqrt{(x + \sqrt{2x - 1})} + \sqrt{(x - \sqrt{2x - 1})} = A`                                       <=>
                  `(\sqrt{(x + \sqrt{2x - 1})} + \sqrt{(x - \sqrt{2x - 1})})^{2} = A^{2}, (A >= 0)`     <=>
                  `2x +  2 * \sqrt{(x + \sqrt{2x - 1})(x - \sqrt{2x - 1})} = A^{2}, (A >= 0)`               <=>

But here, notice that the expression inside square roots is nothing but (`x^2 - 2x + 1`, which is a square expression). Now, again;

                 `2x + 2|x - 1| = A^{2}, (A >= 0, x >= 1/2)`               <=>

(`|x|` is the absolute value of `x`).
The condition `x >= 1/2` is because of `\sqrt{2x - 1}`, where the one inside square-root cant be negative.

Now substitute values for `A`,

(i) `A = 1`, `x + |x - 1| = 1/2`.

Notice that, `|x - 1|` cant be `1 - x`. If it is so, `A^{2}/2 = 1` always, regardless of the actual value of A. So, this yields another condition:

        If `A^{2} / 2 != 1`, then `x >= 1`.

Consider these conditions into the expression (i), and you get `x = 3/4`, which is not according to the conditions. So, (i) does not have a solution.

(ii) `A = 1/2`, `x + |x - 1| = 1/8`.

Here, we get `x = 9/16`, since this is also less than `1`, (ii) also has no solution.

(iii) `A = 2`, `x + |x - 1| = 2`. Here you get, `x = 3/2`, which satisfies all conditions. So, only this combination has solutions for `x`, which is given by `3/2`.

3. Remember `cos(2x) = 2cos^{2}(x) - 1`. And, in a quadratic equation `ax^{2} + bx + c = 0`, the sum of roots is given by `-b/a` and product is given by `c/a`. First, derive formulae for `cos(2x_{1}) + cos(2x_{2}) ` and `cos(2x_{1})cos(2x_{2})` in terms of `cos(x_{1}) + cos(x_{2})` and `cos(x_{1})cos(x_{2})`, where `x_{1}` and `x_{2}` are roots of original equation.

               `cos(2x_{1}) + cos(2x_{2}) = 2cos^{2}(x_{1}) - 1 + 2cos^{2}(x_{2}) - 1`         <=>
                   `= 2(cos(x_{1}) + cos(x_{2}))^{2} - 4cos(x_{1})cos(x_{2}) - 2`.

(Since `x^2 + y^2 = (x + y)^2 - 2xy`).

              `cos(2x_{1})cos(2x_{2}) = (2cos^{2}(x_{1}) - 1) * (2cos^{2}(x_{2}) - 1)`                 <=>
                   `= 4cos^{2}(x_{1}) * cos^{2}(x_{2}) - 2cos^{2}(x_{1}) - 2cos^{2}(x_{2}) + 1`  <=>
                   `= 4(cos(x_{1})cos(x_{2}))^{2} - 2(cos(x_{1}) + cos(x_{2}))^{2} + 4cos(x_{1})cos(x_{2})  + 1`.


            `cos(x_{1}) + cos(x_{2}) = -b/a` and `cos(x_{1})cos(x_{2}) = c/a`, you get the following:

            `-b^{1}/a^{1} = 2 * (-b/a)^{2} - 4 * (c/a) - 2` <=>
                `= (2b^{2} - 4ca -2a^{2})/a^{2}`

            `c^{1}/a^{1} = 4 * (c/a)^{2} - 2 * (-b/a)^{2} + 4 * (c/a) + 1` <=>
                 `= (4c^{2} - 2b^{2} + 4ca + a^{2}) / a^{2}`

So, one can use: `a^{1}  = a^{2}`, `b^{1} = (-2b^{2} + 4ca + 2a^{2})` and `c^{1} = (4c^{2} - 2b^{2} + 4ca + a^{2})`.

The resulting expression is, `a^{2} cos^{2}(2x) + (-2b^{2} + 4ca + 2a^{2}) cos(2x) + (4c^{2} - 2b^{2} + 4ca + a^{2}) = 0`.

When you substitute, `a = 4, b = 2, c = -1`, you get. the roots for original equation:

                      `\alpha = (-1 + \sqrt{5}) / 4, and \beta = (-1 - \sqrt{5}) / 4`.

Surprisingly, the equation in `cos(2x)` also yields the same roots. and in fact, the sane equation when simplified by eliminating common factors between co-efficients.

However, there is no need to be surprised. Because, calculate `cos(2x)` for `\alpha`, it will be `\beta` and `cos(2x)` for `\beta` will be `\alpha`. So, both the equations are right, and compatible. (Its a neat trick substitution).

4. Consider an angle in the right triangle to be `\theta`, where `\theta != 90^{o}`. The median divides the hypotenuse into 2 equal lengthed parts. Since the lengh of hypotenuse is `c`, one side will be `c * cos(\theta)` and another will be `c * sin(\theta)`. The median divides the right triangle into two other triangles. Consider the triangle inside which we have angle `\theta`. Using the cosine rule of lengths of sides, gives the following:

(To brush up: Cosine rule says that, if angle between sides length `a` and `b` is `\theta`, the third side in the triangle is given by `sqrt{a^{2} + b^{2} - 2ab * cos(\theta)}`).

        The geometric mean of sides is, `sqrt{c * cos(\theta) * c * sin(\theta)}`
                     = `c * sqrt{sin(2 * \theta) / 2}`

        Using cosine rule,

       `c^{2} * sin(2 * \theta) / 2 = c^{2} / 4 + c^{2} * cos^{2} (\theta) - 2 * c/2 * c * cos(\theta) * cos(\theta)` <=>
            `c^{2} * sin(2 * \theta) / 2 = c^{2} / 4 + c^{2} * cos^{2} (\theta) - c^{2} * cos^{2} (\theta)`     <=>
            `c^{2} * sin(2 * \theta) / 2 = c^{2} / 4`,

       So, the side length is `c / 2` as well (note that the length of parts of hypotenuse after being divided by the median, are also same as `c/2`). Now, to the angle `\theta`.

             `(sin(2 * \theta)) / 2 = 1/ 4` <=>
             `sin(2 * \theta) = 1/2`       <=>
             `2 * \theta = 30^{o}` or `2 * \theta = 150^{o}` (since `sin(30^{o}) = sin(150^{o}) = 1/2`).
              `\theta = 15^{o}` or `\theta = 75^{o}` (In fact, both angles are complementary in a right angled triangle).

Basically, to construct such a triangle, with any `c`, choose an angle to be `15^{o}` in the right triangle.

5. Now to the last problem, this is a geom problem. I will present anal. geom. solution except for the last part. (Which is too easy in co-ordinate geometry).

    Consider this theorem, well known in circles of (pun is intended), ananlytical geom. If a Jyaa, `AB` in a circle makes an angle `\theta` at the center `C` of the circle, (namely, `<ACB` is `\theta`), then on any point on the circle, to the side of `C`, it makes an angle of `< \theta / 2`). With the help of this theorem, you can easily solve this problem.

        First of all, notice that `<ANM` and `<MNB` are both `45^{o}`, since `<APM` and `<MQB` are `90^{o}` (`P` and `Q` are also centers of circum-circle for a square).

        Consider `<NAM = \theta`. `<NMA = 135^{o} - \theta`. `<NMB = 45^{o} + \theta` and `<NBM = 90^{o} - \theta`. Apply sine rule now, using `NM` which is same in both triangles, `\delta NMA` and `\delta NMB`.

        `(NM)/sin(\theta) = (AM)/sin(45^{o})` and `(NM)/sin(90^{o} - \theta) = (MB)/sin(45^{o})`

This gives, `sin(\theta) = (NM)/(sqrt{2} * AM)` and `cos(\theta) = (NM)/(sqrt{2} * MB)`, which both can be combined to give,

                 `tan(\theta) = (MB)/(AM)`.

But, notice that, `(MB)/(AM) = (MB)/(MC) = (FM)/(AM)`. so, this, along with the facts that `<NAM = \theta` and `<NBM = 90^{o} - \theta`, gives that, `A`, `N` and `F` are co-linear, as well as, `B`, `C` and `N` are co-linear.

(ii) Consider for now, `A` is the origin, and `AB` is the `x` axis. Consider `AM = l_{1}` and `AB = l`. Consider the point `R` at `(l/2, -l/2)`. The angle `<BMR = 135^{o} - \theta` since the opposite angle `<AMN` is of same measure.

        `tan(<BMP) = tan(135^{o} - \theta) = -tan(45^{o} + \theta) = (1 + tan(\theta))/(tan(\theta) - 1)`.

        We also know that.

         `tan(\theta) = (l - l_{1}) / l_{1}`, substituting it.

         `tan(<BMP) = l / (l - 2 * l_{1})` and notice that with mid-point `S` of `AB`, `MS` length = `l/2 - l_{1}`.

          This gives length of `SR = (MS) * tan(<BMP) = (l/2 - l_{1}) * l / (l - 2 * l_{1})`  <=>
                 `SR = (l - 2 * l_{1}) / 2 * l / (l - 2 * l_{1}) = l / 2`

which is irrespective of any `\theta` or equivalently `l_{1}`.

(iii) Using co-ord geom, `P` is `(l_{1} / 2 , l_{1} / 2)` and `Q` is `(l_{1} + (l - l_{1})/2, (l - l_{1})/2)` which is same as, `((l + l_{1})/2, (l - l_{1})/2)`. So, their mid-point is given by `((l + (2 *  l_{1}))/4, l/2)`. So, that means locus is, `y = l/2`.

Monday, October 1, 2012

Cool Math Olympics

    International Math Olympiad problems will be presented along with solutions here in "Cool Math Olympics". Olympiad problems from 1959 (first time Math Olympiad started) to current year will be presented and solved here. First set of Olympiad problems and solutions (one day gap) for 1959 Olympiad is here at "1959 Cool Math Olympics". Olympians, or wannabe Olympians, participate in our "Cool Math Olympics". You performance is much appreciated.

1. Prove that the fraction `(21n+4) / (14n+3)` is irreducible for every natural number `n`. (Hint: What is the definition of GCD? And if `a` and `b` have GCD of `1`, what about `a` and `a - b` or, `b` and `a - b`).
2. For what real values of x is

                `\sqrt{(x + \sqrt{2x - 1})} + \sqrt{(x - \sqrt{2x - 1})} = A`,
given (a) `A =
1` (b) `A = 1/2`, (c) `A = 2`
where only non-negative real numbers are admitted for square roots? (Hint: Too many square roots, looks like it requires raising to power two multiple times, or is it?)
3. Let `a, b, c` be real numbers. Consider the quadratic equation in `cos(x)`:
                `a * cos^{2}(x) + b * cos(x) + c = 0`
Using the numbers `a, b, c` form a quadratic equation in `cos(2x)`, whose roots are the same as those of the original equation. Compare the equations in `cos(x)` and `cos(2x)` for `a = 4, b = 2, c = -1`. (Hint: Use formulas for sum and product of roots, also apply same for `cos(2x)`).
    3a. Bouns question: What happened when you used `a = 4, b = 2, c = -1` and why?
4. Construct a right triangle with given hypotenuse `c` such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle. (Hint: Parameterize by `\theta`).
    4a. What is the angle `\theta` got? Is it dependent on `c`?
    4b. What is the geometric mean? What is it same as?
5. An arbitrary point `M` is selected in the interior of the segment `AB`. The squares `AMCD` and `MBEF` are constructed on the same side of `AB`; with the segments `AM` and `MB` as their respective bases. The circles circumscribed about these squares, with centers `P` and `Q`; intersect at `M` and also at another point `N`. Let `N^{1}` denote the point of intersection of the straight lines `AF` and `BC`.
    (a) Prove that the points `N` and `N^{1}` coincide.
    (b) Prove that the straight lines `MN` pass through a fixed point `S` independent of the choice of `M`.
    (c) Find the locus of the midpoints of the segments `PQ` as `M` varies between `A` and `B`.
(Hint: Use Cartesian Co-ordinates. Like many such problems, choose origin and axes to minimize the paraphernalia).
    5d. Bonus problem. There is a way to solve 5a, using analytical geometry.
    5e. Prove that `AF` and `BC` intersect making an angle of `90^{0}`.
    5f. Prove also using analytical geometry, (use `\theta` where `tan(\theta) = (AM)/(MB)` as well) that there is such a fixed point like `S`.
    5g. Argue that such a fixed point, would be on the perpendicular bisector of `AB` (use symmetry arguments, possibly?)
6. Two planes, `P` and `Q` intersect along the line `p`. The point A is given in the plane `P` and the point `C` in the plane `Q`; neither of these points lies on the straight line `p`. Construct an isosceles trapezoid `ABCD` (with `AB` parallel to `CD`) in which a circle can be inscribed, and with vertices `B` and `D` lying in the planes `P` and `Q` respectively. (Don't know what this problem is about. Whats the idea of this problem, anyway? What is "in which a circle can be inscribed", totally vague).

    1959 Cool Math Olympics complete. Boy, these things take time, Phew. I personally prefer analytical geometry approach to problems than Co-ordinate geom approach, that's just dry, what do you people think? Though these problems are way back in 1959, Olympics solving took time for me, partly because I dealt with these math subjects, long long long time ago (15 years). However, these problems rank as "brilliant" for their formulation, where the results are pretty surprising (Especially 4 and 5). Hats-off to those people who had such great geometric insights. (Spoiler: Pretty much everything in geometry has an interesting locus, or passes through a fixed point I think. What do you people think?)

    People who could crack these (with or without hints) can start telling their friends that they are good enough to do it. Start the celebrations, and the geekery of you telling others about these problems. Next time, with 1960 Cool Math Olympics.