NMTC Bhaskara ( junior level) 9th and 10th Aug 31 2024

From $$x^2 = 1 - x$$ the higher powers can be reduced step by step, giving $$x^3 = 2x - 1$$, $$x^4 = 2 - 3x$$, $$x^5 = 5x - 3$$, $$x^6 = 5 - 8x$$ and $$x^7 = 13x - 8$$. Hence $$\frac{x^7 + 34}{x + 2} = \frac{13x - 8 + 34}{x + 2} = \frac{13(x + 2)}{x + 2} = 13$$.

The minute hand at 38 minutes is at $$38 \times 6^\circ = 228^\circ$$ from the 12 mark. The hour hand is at $$9 \times 30^\circ + 38 \times 0.5^\circ = 270^\circ + 19^\circ = 289^\circ$$. The angle between them is $$289^\circ - 228^\circ = 61^\circ$$.

TO BE FILLED - figure required

Collecting like bases, $$2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$$, that is $$\frac{3}{2} \cdot 2^{2x} = \frac{4}{\sqrt{3}} \cdot 3^{x}$$. This simplifies to $$2^{2x - 3} = 3^{\frac{2x-3}{2}}$$, so $$\left(\frac{2}{\sqrt{3}}\right)^{2x - 3} = 1$$ and $$2x - 3 = 0$$, giving $$x = \frac{3}{2}$$. With $$a = 3$$ and $$b = 2$$, $$\frac{a+b}{a-b} = \frac{5}{1} = 5$$.

If $$x^2 + 1$$ divides the cubic, the other factor must be $$x + a$$, so $$x^3 + ax^2 + bx + c = (x^2 + 1)(x + a) = x^3 + ax^2 + x + a$$. Comparing coefficients gives $$b = 1$$ and $$c = a$$. Since $$a$$ can be any of the 12 allowed values and then $$c$$ is fixed, there are 12 such polynomials.

Pairing the factors, the left side numerator is $$(u + 10)(u + 12)$$ with $$u = x^2 + 7x$$ and the denominator is $$(v + 10)(v + 12)$$ with $$v = x^2 - 7x$$. The equation becomes $$u^2 + 22u = v^2 + 22v$$, that is $$(u - v)(u + v + 22) = 0$$. Since $$u - v = 14x$$ and $$u + v + 22 = 2x^2 + 22 > 0$$, the only real solution is $$x = 0$$.

Squaring gives $$a^2 = 23a + b$$ and $$b^2 = 23b + a$$. Subtracting, $$(a-b)(a+b) = 22(a-b)$$, and as $$a \neq b$$ this gives $$a + b = 22$$. Adding the two equations, $$a^2 + b^2 = 24(a + b) = 528$$, so $$\sqrt{a^2 + b^2 + 48} = \sqrt{576} = 24$$.

Since $$PA = PB$$ and the angle at $$P$$ is $$72^\circ$$, the base angles give $$\angle PAB = \angle PBA = 54^\circ$$. As $$AC$$ is parallel to $$PB$$, alternate angles give $$\angle CAB = \angle ABP = 54^\circ$$, and the tangent-chord angle at $$B$$ together with the parallel lines gives $$\angle ACB = 54^\circ$$ as well, so $$\angle ABC = 180^\circ - 54^\circ - 54^\circ = 72^\circ$$. Since $$ABCD$$ is cyclic, $$\angle CDA = 180^\circ - 72^\circ = 108^\circ$$.

Squaring gives $$19^8 + 19^x = 19^4(19^x + 1)$$, so $$19^8 - 19^4 = 19^x(19^4 - 1)$$, that is $$19^4(19^4 - 1) = 19^x(19^4 - 1)$$ and hence $$x = 4$$. Substituting $$x = 4$$ in the options, only $$3(16) - 11(4) - 4 = 48 - 48 = 0$$ holds.

The general term is $$n(n+3)^2 = n^3 + 6n^2 + 9n$$ for $$n$$ from 1 to 25. Using $$\sum n^3 = 105625$$, $$\sum n^2 = 5525$$ and $$\sum n = 325$$, we get $$S = 105625 + 6(5525) + 9(325) = 141700$$. Hence $$\frac{S}{325} = 436$$.

Taking reciprocals, $$\frac{1}{a_n} = \frac{1}{a_{n-1}} + 2n$$ with $$\frac{1}{a_1} = 2$$, which gives $$\frac{1}{a_n} = n(n+1)$$. So $$a_n = \frac{1}{n} - \frac{1}{n+1}$$ and the sum telescopes to $$1 - \frac{1}{2025} = \frac{2024}{2025}$$.

Taking logarithms to base 10 and writing $$t = \log_{10} x$$, the equation becomes $$(1 + t)t = 1 + t$$, so $$t^2 = 1$$ and $$t = \pm 1$$. Hence the roots are $$\alpha = 10$$ and $$\beta = \frac{1}{10}$$, giving $$\alpha + \frac{1}{\beta} = 10 + 10 = 20$$.

Circles inscribed in a fixed angle and touching one another are similar figures scaled from the vertex, so their radii form a geometric progression. With $$r_1 = 7$$ and $$r_3 = 7k^2 = 28$$ we get $$k = 2$$. Hence the fourth radius is $$r_4 = 7 \times 2^3 = 56$$.

The wrong equation had product of roots $$(-2)(-15) = 30$$, so $$q = 30$$ is correct. The original equation is $$x^2 + 13x + 30 = 0$$ with roots $$\alpha = -3$$ and $$\beta = -10$$. Then $$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{9 + 100}{30} = \frac{109}{30}$$ and the product is 1, so the required equation is $$30x^2 - 109x + 30 = 0$$.

Factorising each bracket, $$1 + xy + x + y = (1 + x)(1 + y)$$ and $$1 - xy + x - y = (1 + x)(1 - y)$$. So the left side is $$(1 + x)^2[(1 + y)^2 - (1 - y)^2] = (1 + x)^2 (4y)$$. Comparing with $$ky(1 + x)^2$$ gives $$k = 4$$.

Since $$x^{10} + 1 = (x^2)^5 + 1$$ and the exponent 5 is odd, $$x^2 + 1$$ divides it exactly with quotient $$x^8 - x^6 + x^4 - x^2 + 1$$. So the non-zero coefficients are $$a = 1$$, $$c = -1$$, $$e = 1$$, $$g = -1$$, $$k = 1$$ and the rest are 0. Raising to the even power 2024 turns each of these five into 1, so the sum is 5.

The four positive roots have sum 4 and product 1, so their arithmetic mean is 1 and their geometric mean is also 1. Equality in the AM-GM inequality forces all four roots to equal 1, so the equation is $$(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$$. Hence $$a = 6$$, $$b = -4$$ and $$a + b = 2$$.

Since $$AB$$ is perpendicular to the diameter at $$B$$, it is a tangent at $$B$$, so the two tangents from $$E$$ give $$EB = ED$$ and hence $$\angle EBD = \angle EDB$$. Because $$BOC$$ is a diameter, $$\angle BDC = 90^\circ$$, so $$\angle ADB = 90^\circ$$ and $$\angle ADE = 90^\circ - \angle EDB = 90^\circ - \angle ABD = \angle DAE$$. Therefore $$EA = ED = EB$$, making $$E$$ the midpoint of $$AB$$, so $$AB = 2(AE)$$ and $$k = 2$$.

In any triangle $$\tan A + \tan B + \tan C = \tan A \tan B \tan C$$, so writing the tangents as $$t, 2t, 3t$$ gives $$6t = 6t^3$$ and $$t = 1$$. Hence $$\tan B = 2$$ and $$\tan C = 3$$, so $$\sin B = \frac{2}{\sqrt{5}}$$ and $$\sin C = \frac{3}{\sqrt{10}}$$. By the sine rule $$\frac{AC}{AB} = \frac{\sin B}{\sin C} = \frac{2\sqrt{2}}{3}$$, so $$p + q + r = 2 + 2 + 3 = 7$$.

Dividing successively by 5, 6 and 7 with remainders 3, 2 and 2 means the number has the form $$210k + 73$$. As the number was meant to be divided by 120, it exceeds 120, and the first such value is $$210 + 73 = 283$$. Since $$283 = 2 \times 120 + 43$$, the remainder is 43.

If a number leaves the same remainder on dividing 30, 53 and 99, it must divide each of the differences $$53 - 30 = 23$$, $$99 - 53 = 46$$ and $$99 - 30 = 69$$. The greatest such number is their HCF, which is 23.

Putting $$y = x + 1$$, so that $$x = y - 1$$, gives $$f(y) = (y-1)^2 - 3(y-1) + 2 = y^2 - 5y + 6$$. The roots of $$y^2 - 5y + 6 = 0$$ are 2 and 3, so $$\alpha^2 + \beta^2 = 4 + 9 = 13$$.

The largest cylinder has radius $$\frac{a}{2}$$ and height $$a$$, so its volume is $$\pi \frac{a^3}{4} = \frac{22}{7} \times \frac{a^3}{4} = \frac{11a^3}{14}$$. The remaining solid has volume $$a^3 - \frac{11a^3}{14} = \frac{3a^3}{14}$$, so $$k = \frac{3}{14}$$ and $$p + q = 3 + 14 = 17$$.

Since $$x = 1$$ and $$x = -2$$ are roots, the polynomial factorises as $$(x - 1)(x + 2)(5x^2 + 4x + 4)$$. The quadratic $$5x^2 + 4x + 4$$ has discriminant $$16 - 80 < 0$$, so it is the irreducible factor. Hence $$a = 5$$, $$b = 4$$, $$c = 4$$ and $$a^2 + b^2 - c^2 = 25 + 16 - 16 = 25$$.

The square has side 6, so with O as origin the vertices are $$A(-6, 0)$$, $$B(-6, 6)$$ and $$C(0, 6)$$, and since $$B$$ lies on the semicircle the radius is $$OB = 6\sqrt{2}$$. The line joining $$Q(6\sqrt{2}, 0)$$ to $$C$$ meets the arc again at $$D(-2\sqrt{2}, 8)$$, so $$QD = \sqrt{(8\sqrt{2})^2 + 8^2} = \sqrt{192} = 8\sqrt{3}$$. Hence $$x\sqrt{3} = 8\sqrt{3} \times \sqrt{3} = 24$$.

The bracket equals $$1 + \frac{1}{4}\left(\frac{a}{b} + \frac{b}{a} - 2\right) = \frac{4ab + a^2 + b^2 - 2ab}{4ab} = \frac{(a+b)^2}{4ab}$$. Its square root is $$\frac{a+b}{2\sqrt{ab}}$$, so the whole expression is $$\frac{2\sqrt{ab}}{a+b} \times \frac{a+b}{2\sqrt{ab}} = 1$$ whatever the values of $$a$$ and $$b$$.

With first term $$A$$ and ratio $$r$$, the condition $$\frac{A}{1-r} = \frac{1}{8} \cdot \frac{A^2}{1-r^2}$$ gives $$A = 8(1 + r)$$, and $$Ar = 6$$ then gives $$4r^2 + 4r - 3 = 0$$, so $$r = \frac{1}{2}$$ and $$A = 12$$. The first and fourth terms add to $$12 + \frac{12}{8} = \frac{27}{2}$$, so $$\left[\frac{p}{q}\right] = \left[13.5\right] = 13$$.

Since $$\frac{\sqrt{10}}{10} = 10^{-1/2}$$, the expression is $$10^{-\frac{1}{2}(\log_{10} 9 - 2)} = 10^{-\frac{1}{2}\log_{10} 9} \times 10 = \frac{10}{\sqrt{9}} = \frac{10}{3}$$. Hence $$a = 10$$, $$b = 3$$ and $$a - b = 7$$.

Let the side of the square be $$2r$$ and $$DE = t$$. Equal tangents give $$BE = BA + DE = 2r + t$$, and the right triangle $$BCE$$ gives $$(2r + t)^2 = (2r)^2 + (2r - t)^2$$, which simplifies to $$t = \frac{r}{2}$$. Then the area is $$\frac{1}{2} \times 2r \times \left(2r - \frac{r}{2}\right) = \frac{3r^2}{2} = 216$$, so $$r^2 = 144$$ and $$r = 12$$.

Every power appearing in $$f$$ is odd, so $$f$$ is an odd function and $$f(4) = -f(-4) = -18$$, giving $$|f(4)| = 18$$. Since $$|2\cos x|$$ can be at most 2, the maximum value of the sum is $$18 + 2 = 20$$.