MaxWong

Answered here: https://web2.0calc.com/questions/number-theory_23228

Let x be the minimum length of CQ. Note that by Power of Points formula, AQ * BQ = CQ * DQ.

\(AQ \cdot BQ = CQ \cdot DQ\\ 16 \cdot 12 = x \cdot (36 - x)\\ x^2 - 36x + 192 = 0\\ x = 18 - 2 \sqrt {33}\text{ or }x = 18 + 2\sqrt {33}\)

The smaller one is 18 - 2 * sqrt(33), and that's the answer.

Answered here: https://web2.0calc.com/questions/please-help_87485#r1

Notice that for any 2 consecutive terms in the product, the numerator of the first fraction always cross out with the denominator of the second fraction. That leaves us with the numerator of the last fraction, divided by the denominator of the first fraction, which is (n + 20)/n. If you are attempting the solve the equation, you should try to multiply both sides by n so that the equation becomes linear.

You know that 

\(a_1 = 48 - 3\times1 = 45\\ d = a_2 - a_1 = (48 - 3(2)) - (48 - 3(1)) = -3\\ \)

where d is the common difference of the sequence.

The sum of A.S. formula states \(a_1 + a_2 + \cdots + a_n = \dfrac n2 \left(2a_1 + (n - 1)(a_2 - a_1)\right) = \dfrac n2 (2a_1 + (n - 1)d)\) if a_n is an arithmetic sequence. Therefore, the equation \(a_1 + a_2 + a_3 + \cdots + a_k = 330\) really says \(\dfrac k2 (2(45) + (k - 1)(-3)) = 330\). This becomes a quadratic equation. Can you solve for the value of k now?

Hint: Expand and use quadratic formula.

You can let \(f(x) = x^4 + ax^3 + bx^2 + cx + d\) be the polynomial. You can then use the given information to find the missing coefficients.

Substituting x = -1, 2, -3, 4 respectively gives \(\begin{cases}1-a+b-c+d = -1\\16 + 8a + 4b + 2c + d = -4\\81 - 27a + 9b - 3c + d = 9\\256 + 64a + 16b + 4c + d = -16\end{cases}\).

Solving gives: 

So f(1) = \(1 + \left(-\dfrac{79}{35}\right) + \left(-\dfrac{89}7\right) + \dfrac{432}{35} + \dfrac{768}{35}\) = 711/35.

This is the diagram.

Note that \(\angle BPC = 180^\circ - 116^\circ = 64^\circ\).

Since BC = CP, \(\angle PBC = 64^\circ\).

Considering the interior angle sum of \(\triangle BPC\) gives \(\angle PCB = 52^\circ\).

Since AB = AC, \(\angle ACB = \angle ABC = \angle PBC = 64^\circ\).

Therefore, \(\angle ACP = \angle ACB - \angle PCB = 64^\circ - 52^\circ = 12^\circ\).

Consider the prime factorization of n. Suppose that \(n = 2^x 3^y 5^z\). (It does not have any other prime factors because n is the smallest one)

Then \(2n = 2^{x + 1} 3^y 5^z\), \(3n = 2^{x} 3^{y + 1} 5^z\), and \(5n = 2^x 3^y 5^{z + 1}\).

2n being a perfect square means all exponents are divisible by 2, i.e., x is odd, y and z are even.

3n being a perfect cube means all exponents are divisible by 3, i.e., x, y + 1, and z are divisible by 3.

5n being a perfect fifth power means all exponents are divisible by 5, i.e., x, y, and z + 1 are divisible by 5.

The smallest odd value of x that is divisible by 3 and 5 is 15.

The smallest even value of y such that y + 1 is divisible by 3 and y is divisible by 5 is 20.

The smallest even value of z such that z is divisible by 3 and z + 1 is divisible by 5 is 24.

Therefore, \(n = 2^{15}3^{20}5^{24}\) is the required answer.

Since the coefficients are rational and the roots are irrational, the roots are \(x = 1\pm \sqrt 2\) and \(x = 2\pm \sqrt 7\).

The quartic polynomial required is 

\(\quad(x - (1 + \sqrt 2))(x - (1 - \sqrt 2))(x - (2 + \sqrt 7))(x - (2 - \sqrt 7))\\ = (x^2 - 2x - 1)(x^2 - 4x - 3)\)

Can you expand that on your own?

Let \(A = \begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\), you are given \(\det(A)\) = -24, and you are to find \(\det(A^{\top})\). 

Combine this with the formula I stated. Is this clearer to you now?