[b]p1.[/b] The expression $3 + 2\sqrt2$ can be represented as a perfect square: $3 +\sqrt2 = (1 + \sqrt2)^2$.
(a) Represent $29 - 12\sqrt5$ as a prefect square.
(b) Represent $10 - 6\sqrt3$ as a prefect cube.


[b]p2.[/b] Find all values of the parameter $c$ for which the following system of equations has no solutions.
$$x+cy = 1$$
$$cx+9y = 3$$


[b]p3.[/b] The equation $y = x^2 + 2ax + a$ represents a parabola for all real values of $a$.
(a) Prove hat each of these parabolas pass through a common point and determine the coordinates of this point.
(b) The vertices of the parabolas lie on a curve. Prove that this curve is a parabola and find its equation.


[b]p4.[/b] Miranda is a $10$th grade student who is very good in mathematics. In fact she just completed an advanced algebra class and received a grade of A+. Miranda has five sisters, Cathy, Stella, Eva, Lucinda, and Dorothea. Miranda made up a problem involving the ages of the six girls and dared Cathy to solve it.
Miranda said: “The sum of our ages is five times my age. (By ’age’ throughout this problem is meant ’age in years’.) When Stella is three times my present age, the sum of my age and Dorothea’s will be equal to the sum of the present ages of the five of us; Eva’s age will be three times her present age; and Lucinda’s age will be twice Stella’s present age, plus one year. How old are Stella and Miranda?”
“Well, Miranda, could you tell me something else?”
“Sure”, said Miranda, “my age is an odd number”.


[b]p5.[/b] Cities $A,B,C$ and $D$ are located in vertices of a square with the area $10, 000$ square miles. There is a straight-line highway passing through the center of a square. Find the sum of squares of the distances from the cities of to the highway.
[img]https://cdn.artofproblemsolving.com/attachments/b/4/1f53d81d3bc2a465387ff64de15f7da0949f69.png[/img]


[b]p6.[/b] (a) Among three similar coins there is one counterfeit. It is not known whether the counterfeit coin is lighter or heavier than a genuine one (all genuine coins weight the same). Using two weightings on a pan balance, how can the counterfeit be identified and in process determined to be lighter or heavier than a genuine coin?
(b) There is one counterfeit coin among $12$ similar coins. It is not known whether the counterfeit coin is lighter or heavier than a genuine one. Using three weightings on a pan balance, how can the counterfeit be identified and in process determined to be lighter or heavier than a genuine coin?


PS. You should use hide for answers.
### Problem 1
#### Part (a)
1. We need to represent \( 29 - 12\sqrt{5} \) as a perfect square.
2. Assume \( 29 - 12\sqrt{5} = (a - b\sqrt{5})^2 \).
3. Expanding the right-hand side, we get:
   \[
   (a - b\sqrt{5})^2 = a^2 - 2ab\sqrt{5} + 5b^2
   \]
4. Equate the rational and irrational parts:
   \[
   a^2 + 5b^2 = 29 \quad \text{and} \quad -2ab\sqrt{5} = -12\sqrt{5}
   \]
5. From the irrational part, we get:
   \[
   2ab = 12 \implies ab = 6
   \]
6. Substitute \( b = \frac{6}{a} \) into the rational part:
   \[
   a^2 + 5\left(\frac{6}{a}\right)^2 = 29 \implies a^2 + \frac{180}{a^2} = 29
   \]
7. Multiply through by \( a^2 \) to clear the fraction:
   \[
   a^4 - 29a^2 + 180 = 0
   \]
8. Let \( u = a^2 \), then the equation becomes:
   \[
   u^2 - 29u + 180 = 0
   \]
9. Solve the quadratic equation using the quadratic formula:
   \[
   u = \frac{29 \pm \sqrt{29^2 - 4 \cdot 180}}{2} = \frac{29 \pm \sqrt{841}}{2} = \frac{29 \pm 29}{2}
   \]
10. This gives \( u = 29 \) or \( u = 0 \). Since \( a \neq 0 \), we have:
    \[
    a^2 = 20 \implies a = \sqrt{20} = 2\sqrt{5}
    \]
11. Then \( b = \frac{6}{2\sqrt{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \).
12. Therefore, \( 29 - 12\sqrt{5} = (2\sqrt{5} - \frac{3\sqrt{5}}{5})^2 \).

#### Part (b)
1. We need to represent \( 10 - 6\sqrt{3} \) as a perfect cube.
2. Assume \( 10 - 6\sqrt{3} = (a - b\sqrt{3})^3 \).
3. Expanding the right-hand side, we get:
   \[
   (a - b\sqrt{3})^3 = a^3 - 3a^2b\sqrt{3} + 3ab^2(3) - b^3(3\sqrt{3})
   \]
4. Equate the rational and irrational parts:
   \[
   a^3 + 9ab^2 = 10 \quad \text{and} \quad -3a^2b - 3b^3\sqrt{3} = -6\sqrt{3}
   \]
5. From the irrational part, we get:
   \[
   -3a^2b - 3b^3 = -6 \implies a^2b + b^3 = 2
   \]
6. Substitute \( b = \frac{2 - a^2b}{b^3} \) into the rational part:
   \[
   a^3 + 9a\left(\frac{2 - a^2b}{b^3}\right)^2 = 10
   \]
7. This equation is complex and does not simplify easily, indicating that \( 10 - 6\sqrt{3} \) may not be representable as a perfect cube.

### Problem 2
1. The system of equations is:
   \[
   x + cy = 1
   \]
   \[
   cx + 9y = 3
   \]
2. To find when the system has no solutions, we need to check the determinant of the coefficient matrix:
   \[
   \begin{vmatrix}
   1 & c \\
   c & 9
   \end{vmatrix} = 1 \cdot 9 - c \cdot c = 9 - c^2
   \]
3. The system has no solutions if the determinant is zero:
   \[
   9 - c^2 = 0 \implies c^2 = 9 \implies c = \pm 3
   \]

### Problem 3
#### Part (a)
1. The equation of the parabola is:
   \[
   y = x^2 + 2ax + a
   \]
2. To find the common point, substitute \( x = 0 \):
   \[
   y = 0^2 + 2a(0) + a = a
   \]
3. Therefore, the common point is:
   \[
   (0, a)
   \]

#### Part (b)
1. The vertex of the parabola \( y = x^2 + 2ax + a \) is:
   \[
   \left(-a, a - a^2\right)
   \]
2. The vertices lie on the curve:
   \[
   y = -x^2 - x
   \]
3. This is a parabola.

### Problem 4
1. Let Miranda's age be \( M \).
2. The sum of their ages is \( 5M \).
3. When Stella is \( 3M \), the sum of Miranda's and Dorothea's ages is:
   \[
   M + D = 5M - (M + S + E + L)
   \]
4. Eva's age will be \( 3E \).
5. Lucinda's age will be \( 2S + 1 \).
6. Miranda's age is an odd number.

### Problem 5
1. The area of the square is \( 10,000 \) square miles, so the side length is:
   \[
   \sqrt{10,000} = 100 \text{ miles}
   \]
2. The distance from each city to the highway is:
   \[
   \frac{100}{2} = 50 \text{ miles}
   \]
3. The sum of the squares of the distances is:
   \[
   4 \times 50^2 = 4 \times 2500 = 10,000
   \]

### Problem 6
#### Part (a)
1. Weigh two coins against each other.
2. If they balance, the third coin is counterfeit.
3. If they do not balance, the lighter or heavier coin is counterfeit.

#### Part (b)
1. Divide the 12 coins into three groups of four.
2. Weigh two groups against each other.
3. If they balance, the counterfeit is in the third group.
4. If they do not balance, the counterfeit is in the lighter or heavier group.
5. Repeat the process with the remaining coins.

The final answer is \( \boxed{ y = -x^2 - x } \)