Archive for May, 2023

MIT Integration Bee 2023 Semi-final Problem 1

Posted: May 19, 2023 in Mathematics
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I was browsing the net couple days ago and saw this integral. At first it looks quite difficult. I have tried numerous different substitutions and they don’t work. And later I tried using complex number and it seems to work. So, here is my solution.

Solution:

By Euler’s formula, cosine function is the real part of a complex number.

e^{i\theta} = \cos \theta + i \sin \theta

Hence,

\displaystyle \int e^{\cos x} \cos(2x + \sin x) \text{ d}x = \Re \int e^{\cos x} e^{i(2x+\sin x)} \text{ d}x

Re-arrange the exponents we’ll get

\displaystyle = \Re \int e^{\cos x + i\sin x} e^{2ix} \text{ d}x = \Re \int e^{e^{ix}} e^{2ix}\text{ d}x

Now, since

\displaystyle \frac{\text{d}}{\text{d}x} e^{e^{ix}}=ie^{e^{ix}}e^{ix}

We can use integration by parts,

\displaystyle = \Re \; \frac{1}{i} \int e^{ix} \text{ d}e^{e^{ix}}

\displaystyle = \Re \; \frac{1}{i}\left(e^{ix}e^{e^{ix}} - \int e^{e^{ix}}\cdot i e^{ix} \text{ d}x \right)

\displaystyle = \Re \; \frac{1}{i}\left(e^{ix}e^{e^{ix}} - e^{e^{ix}}\right) + C

\displaystyle = \Re \; \frac{1}{i} e^{e^{ix}} \left(e^{ix}-1\right) + C

\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} (\cos x + i\sin x - 1) + C

\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} (\cos x - 1 + i\sin x) + C

By using the double angle identity of sine and cosine,

\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} \left(-2\sin^2 \frac{x}{2} + i2\sin \frac{x}{2} \cos\frac{x}{2}\right) + C

\displaystyle = \Re \; e^{\cos x + i\sin x} \left(2\sin\frac{x}{2}\cos\frac{x}{2} + i\left(2\sin^2 \frac{x}{2}\right)\right) + C

\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x + i\sin x} \left(\cos\frac{x}{2} + i\sin \frac{x}{2}\right) + C

\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x + i\sin x} e^{i\frac{x}{2}} + C

\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x} e^{i\left(\frac{x}{2}+\sin x\right)} + C

\displaystyle = 2 e^{\cos x} \sin\frac{x}{2} \cos\left(\frac{x}{2} + \sin x\right) + C

DSE Math 2023 Paper 1 Answer

Posted: May 2, 2023 in Mathematics
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5h-15=hk+k^2

5h-hk=k^2+15

h(5-k)=k^2+15

h=\dfrac{k^2+15}{5-k}

= \dfrac{y(x^7 y^9)^6}{x^8} = \dfrac{yx^{42}y^{54}}{x^8} = x^{34}y^{55}

The actual mass of a packet of cheese is [215 g, 225 g), hence the total mass of 250 packets is within the interval [53750 g, 56250 g) or [53.75 kg, 56.25 kg). 53.6 kg correct to the nearest 0.1 kg lies outside the interval of [53.75 kg, 56.25 kg). Therefore the claim is false.

(a)

3x+2>\dfrac{4x-5}{2}

6x+4>4x-5

2x>-9

x>-\dfrac{9}{2}

and

3x-2<7

3x<9

x<3

take the intersection of these two intervals,

-\dfrac92<x<3

(b)

-4.5 < x < 3 implies negative integer x could be -4, -3, … , -1, a total of 4 negative integers.

Let F be the number of female passengers and M be the number of male passengers. The first sentence implies

F = 1.4M

Now, 24 female passengers leave.

M = 1.4(F-24)

Substitute the first equation into the second yields,

M = 1.4(1.4M-24)

Since 1.4=\frac{14}{10} = \frac75, then

\displaystyle M = \frac75\left(\frac75M-24\right)

25M=7(7M-120)

25M=49M-840

840=24M

M=35

There are 35 male passengers.

(a) 7a = 6b implies a:b = 6:7

\dfrac{4a-3c}{2b-c}=9

4a-3c=18b-9c

6c=18b-4a

Since 6b=7a, so 18b=21a.

6c=21a-4a

17a=6c

\displaystyle\frac{a}{c} = \frac{6}{17}

Therefore a:b:c = 6:7:17.

(b)

\displaystyle \frac{5a+8b}{7b+3c} = \frac{\frac{5a}{b}+8}{7+\frac{3c}{b}} = \frac{5(\frac67)+8}{7+3(\frac{17}{7})} = \frac{30+56}{49+51}=\frac{86}{100}=\frac{43}{50}

By inscribed angle theorem,

\angle PRQ = \angle PSQ = 41^\circ

By exterior angle theorem,

\angle PTQ = \angle RQS + \angle PRQ

hence, \angle RQS = 68^\circ - 41^\circ = 27^\circ

By inscribed angle of diameter,

\angle PQS + \angle RQS = \angle PQR = 90^\circ

\angle PQS + 27^\circ = 90^\circ

\therefore \angle PQS = 63^\circ

(a)

Given that AC // DB. By alternate angles are equal, angle C and angle D are equal. And angle A and angle B are equal. Then by third angles of the triangle are equal, angle AEC and angle BED are equal. Therefore,

\triangle ACE \sim \triangle BDE

(b)

Since the corresponding sides of the two similar triangles are proportional,

\displaystyle \frac{AE}{BE} = \frac{AC}{BD} = \frac{10}{15} = \frac{2}{3}

\therefore AE = 20\cdot \frac{2}{5} = 8 \text{ cm}

\displaystyle AE^2 + CE^2 = 8^2 + 7^2 = 113 \neq 10^2 = BD^2

Therefore triangle ACE is not a right triangle and hence triangle BDE is not a right triangle as well because they are similar.

(a) Range = 27 hence

a = 49 - 27 -20 = 2

\displaystyle \text{mean} = \frac{20(7) + 2 + 5 + 5 + 6 + 6 + 8 + 8 + 30(8) + 3 + 3 + 3 + 4 + 5 + 5 + 9 + 9 + 40(9) + 1 + 4 + 4 + 5 + 6 + 7 + 7 + 9}{7+8+9}

\displaystyle = \frac{140 + 40 + 240 + 41 + 360 + 43}{24} = \frac{864}{24} = \frac{108}{3} = 36

\text{mode} = 33

(b) 14/24 = 7/12

(a) Line Γ is the perpendicular bisector of line segment AB.

(b) Γ: y = -3x + 12, hence

(i)

m_{\Gamma} = -3

hence the perpendicular slope is

m_{AB} = \dfrac{1}{3}

\displaystyle \ell_{AB}: \quad y+4=\frac{1}{3}(x-2)

\displaystyle y = \frac{x}{3} - \frac{14}{3}

or

\displaystyle x-3y-14=0

(ii) Center of the circle is the intersection of the two lines.

3x+y-12=0 \quad\text{and}\quad x-3y-14=0

Substitute x = 3y + 14 into 3x + y – 12 = 0,

3(3y+14)+y-12=0

9y + 42 + y - 12 = 0

10y + 30 = 0

y = -3

x = 3(-3)+14 = 5

\therefore \text{Center} = (5,-3)

r^2 = (5-2)^2 + (-3-(-4))^2 = 9 + 1 = 10

\therefore (x-5)^2+(y+3)^2=10

a) mean = 2, hence

\displaystyle \frac{8+2\cdot 5 + 3n + 4}{8+5+n+1}=2

3n+22=2(n+14)

n = 6

So, total number of data = 8 + 5 + 6 + 1 = 20.

Q1 = 1 and Q3 = 3, and hence IQR = 3 – 1 = 2.

\displaystyle Var(X) = E(X^2)-E^2(X) = \frac{8(1^2)+5(2^2)+6(3^2)+1(4^2)}{20}-2^2

\displaystyle = \frac{8+20+54+16}{20} -4 = \frac{98}{20} = 4.9-4 = 0.9

(b) The range would change if the student who has 4 calculators is withdrawn from the class. But this is impossible because in order for the mean remains unchanged, the other withdrawn student must have zero calculator but no one has zero calculator.

Given that f(x) = a + bx^2 for some real numbers a and b such that f(10) = 62 and f(15) = 122.

(a)

f(10) = a + 100b = 62 \quad\text{and}\quad f(15) = a + 225b = 122

Subtract the second equation by the first yields,

125b = 60 \quad \therefore b = \dfrac{12}{25}

\therefore a = 62 - 100\cdot\dfrac{12}{25} = 62 - 48 = 14

\therefore f(x) = 14 + \dfrac{12}{25}x^2

\therefore f(5) = 14 + \dfrac{12}{25}\cdot 5^2 = 14 + 12 = 26

(b)

u = f(0) = 14 \quad\text{and}\quad v = f(5) = 14 + \dfrac{12}{25}\cdot 5^2 = 26

Since angle UWV = 90 degrees, UV is diameter of the circle. Hence the circumference is

\displaystyle \pi \sqrt{(5-0)^2 + (26-14)^2} = 13\pi

(a) deg(h) = 4, deg(g) = 3, hence the degree of the quotient function = deg(q) = 1. Since remainder is the same as the quotient, r(x) = q(x). Hence,

h(x) = g(x)q(x) + q(x)

Let q(x) = 3x+k, for some constant k.

h(x)=g(x)(3x+k)+(3x+k) = (3x+k)(g(x)+1) = (3x+k)(x^3+5x^2-12)

= 3x^4+15x^3-36x+kx^3+5kx^2-12k

\therefore 3x^4+ax^3-16x^2+bx+c = 3x^4+(k+15)x^3+(5k-36)x^2-12kx

hence by comparing the coefficients of the equation,

c=0, \quad 5k-36=-16 \text{ or } k=4, \quad b=5k=20, \quad a=15

Therefore the quotient function is 3k + 4.

(b)

h(x) = (3x+4)(g(x)+1) = 0

h(x) = (3x+4)(x^3+5x^2-12x) = 0

x(3x+4)(x^2+5x-12) = 0

x=0, \quad x=-\dfrac{4}{3}

For x^2+5x-12 = 0, \Delta = 25 - 4(-12) = 25+48=73\neq 0 and 73 isn’t a perfect square.

So x^2+5x-12 = 0 has no rational root.

Therefore, h(x)=0 has 2 rational roots.

(a) Since the curved surface area of the cone is πrs where s is the slant height of the cone, then

\pi rs = 700 \pi

14s = 700

s = 50

h=\sqrt{s^2-14^2} = \sqrt{2500-196} = \sqrt{2304} = 48

Therefore the height is 48 cm.

(b) (i)

Since the two triangles in the figure are similar, the ratio of the curved surface area of X and Y is

\displaystyle \frac{h_1^2}{h_2^2} = \frac{1}{1+15} = \frac{1}{16}

\displaystyle \Rightarrow \frac{h_1}{h_2} = \frac{1}{4}

\displaystyle \Rightarrow \frac{h_1^3}{h_2^3} = \frac{1}{64}

Therefore,

\displaystyle \text{Volume of Y} = (\text{Volume of X and Y}) \cdot \frac{64-1}{64}

\displaystyle = \frac{\pi(14^2)(48)}{3}\cdot\frac{63}{64} = 3087 \pi \text{ cm}^3

(ii)

\displaystyle V_{sphere} = \frac{4\pi r^3}{3}

hence,

2\left(\dfrac{4\pi r^3}{3}\right) = 3087

2^3 \cdot r^3 = 3^3 \cdot 7^3

r=\dfrac{21}{2}

Therefore, the diameter of each sphere is 21 cm.

(a)

\displaystyle P(\text{Two chosen balls = Red}) = \frac{5}{9}\cdot\frac{4}{8}=\frac{5}{18}

(b)

Case 1: Drawing 3 R from 10R in the bag,

\displaystyle \frac{5}{18}

Case 2: Drawing 3R from 9R1B in the bag.

\displaystyle \frac{\binom{5}{1} \binom{4}{1}}{\binom{9}{2}}\cdot\frac{\binom{9}{3}}{\binom{10}{3}} = \frac{7}{18}

Case 3: Drawing 3R from 8R2B in the bag.

\displaystyle \frac{\binom{5}{0} \binom{4}{2}}{\binom{9}{2}}\cdot\frac{\binom{8}{3}}{\binom{10}{3}} = \frac{7}{90}

Therefore, the probability that the three chosen balls are of the same color is

\displaystyle \frac{5}{18} + \frac{7}{18} + \frac{7}{90} = \frac{67}{90}

Since p and 5p are the roots of the given equation, by sum of roots and product of roots,

6p = -a \quad\text{and}\quad 5p^2=b

then,

\displaystyle p=-\frac{a}{6} \Rightarrow p^2=\frac{a^2}{36}

Substitute this into 5p^2=b, then

\displaystyle \frac{5a^2}{36} = b

\therefore 5a^2 = 36b

(b) Since OQ : QR = 1 : 4, by similar triangles,

\displaystyle \frac{x_1}{x_2} = \frac{OQ'}{OR'} = \frac{OQ}{OR} = \frac{1}{1+4} = \frac{1}{5}

\therefore x_2 = 5x_1

hence Q(x_1, mx_1) and R(5x_1, 5mx_1)

Plug y=mx into x^2+y^2-6x-12y+20=0, then

x^2+m^2x^2-6x-12mx+20=0

(m^2+1)x^2 - (12m+6)x + 20 = 0

\displaystyle x^2 - \frac{12m+6}{m^2+1}x + \frac{20}{m^2+1} = 0 \qquad (*)

Since x_1 and 5x_1 are roots of (*), hence by part (a),

\displaystyle 5\left(-\frac{12m+6}{m^2+1}\right)^2 = 36\left(\frac{20}{m^2+1}\right)

(2m+1)^2 = 4(m^2+1)

4m^2+4m+1 = 4m^2+4

4m=3

m=\dfrac{3}{4}

(a) By sine law,

\displaystyle \frac{\sin{\angle XWY}}{5} = \frac{\sin{70^\circ}}{6}

\displaystyle\therefore \angle XWY = \sin^{-1} \frac{5\sin{70^\circ}}{6} = 51.54318937...^\circ = 51.5^\circ \text{ (3 sig fig)}

(b) Project Z onto triangle WXY and label it as O. Since slant side = s = WZ = XZ = YZ, we have OX = OY = OW = r = circumradius of triangle WXY. By part(a),

\displaystyle \angle XWY = \sin^{-1} \frac{5\sin{70^\circ}}{6}

Since O is center of the circumcircle,

\angle XOY = 2 \angle XWY

Obviously \triangle XOM \cong \triangle YOM, so

\displaystyle \angle XOM = \frac{1}{2} \angle XOY = \angle XWY= \sin^{-1} \frac{5\sin{70^\circ}}{6}

Triangle WOZ is a 30-60-90 special triangle, so

h = \dfrac{r}{\sqrt3}

In triangle XOM,

\displaystyle \frac{5}{2r} = \sin \angle XOM = \sin \angle XWY

\displaystyle \frac{5}{2r}=\frac{5\sin{70^\circ}}{6}

\displaystyle \therefore r = \frac{3}{\sin{70^\circ}}

\displaystyle h = \frac{3}{\sqrt3 \sin 70^\circ} = \frac{\sqrt3}{\sin70^\circ}

\displaystyle a^2 + \left(\frac52\right)^2 = \left(\frac{3}{\sin70^\circ}\right)^2

\displaystyle a^2 = \frac{9}{\sin^2 70^\circ} - \frac{25}{4} = \frac{36-25\sin^2 70^\circ}{4\sin^2 70^\circ}

\displaystyle a = \frac{\sqrt{36-25\sin^2 70^\circ}}{2\sin70^\circ}

Let θ be the angle between triangles WXY and XYZ.

\displaystyle \theta = \tan^{-1} \frac{h}{a} = \tan^{-1} \frac{\frac{\sqrt3}{\sin70^\circ}}{\frac{\sqrt{36-25\sin^2 70^\circ}}{2\sin70^\circ}} = \tan^{-1}\frac{2\sqrt3}{\sqrt{36-25\sin^2 70^\circ}}

= 42.87142855...^\circ < 45^\circ

Therefore the angle between triangles WXY and XYZ does not exceed 45 degrees.

(a) Suppose α, 7, β form a geometric sequence, then

\displaystyle \frac{7}{\alpha} = \frac{\beta}{7} \Rightarrow \alpha\beta=49 \Rightarrow \alpha = \frac{49}{\beta}

\displaystyle \log_7 \alpha = \log_7\frac{49}{\beta} = \log_7 49 - \log_7 \beta = 2-\log_7 \beta

(b) Let x = \log_7\beta, then the given arithmetic sequence becomes

\displaystyle \frac{2-x}{x}, x, \frac{x}{2-x}

Let d be the common difference,

\displaystyle d = x-\frac{2-x}{x} = \frac{x}{2-x}-x

\displaystyle 2x = \frac{x}{2-x} + \frac{2-x}{x} = \frac{x^2+(2-x)^2}{x(2-x)} = \frac{2x^2-4x+4}{x(2-x)}

\displaystyle x^2(2-x) = x^2-2x+2

\displaystyle 2x^2-x^3= x^2-2x+2

\displaystyle x^3 - x^2 - 2x + 2 = 0

x^2(x-1)-2(x-1)=0

x=1 \quad\text{or}\quad x=\pm\sqrt2

Case x = 1,

\log_7\beta = 1 \Rightarrow \beta = 7

\log_7\alpha = 2-1 = 1 \Rightarrow \alpha = 7

This contradicts with 1 < \alpha < \beta. So reject.

Case x=\sqrt2,

\log_7\beta = \sqrt2

\beta = 7^{\sqrt2} = 15.67289...

\log_7\alpha = 2-\sqrt2

\alpha = 7^{2-\sqrt2} = 3.126417...

\displaystyle d = x - \frac{2-x}{x} = \sqrt2 - \frac{2-\sqrt2}{\sqrt2} = \sqrt2 - \sqrt2 + 1 = 1

Case x=-\sqrt2,

\log_7\beta = -\sqrt2

\beta = 7^{-\sqrt2} = \dfrac{1}{7^{\sqrt2}}

\alpha = \dfrac{49}{\beta} = 49\cdot 7^{\sqrt2} = 7^{2+\sqrt2} > \beta

hence, contradict with the fact that \alpha < \beta, so we reject.

Therefore the common difference is 1.

P(50, 0), Q(32, t), where t > 0. Let R(x, y). Since Q is midpoint of PR, hence

\displaystyle \frac{x+50}{2}=32 \Rightarrow x=14

\displaystyle \frac{y-0}{2} = t \Rightarrow y=2t

\therefore R(14, 2t)

(a) Suppose that G is circumcenter. Let M be midpoint of OP. M(25, 0), so G(25, y).

\displaystyle m_{PR}=\frac{0-2t}{50-14}=\frac{-2t}{36}=-\frac{t}{18} \rightarrow m_{GQ}=\frac{18}{t}

\displaystyle m_GQ=\frac{y-t}{25-32}=\frac{y-t}{-7}=\frac{18}{t} \Rightarrow y=\frac{t^2-126}{t}

\displaystyle \therefore G\left(25, \frac{t^2-126}{t}\right)

Let H(14, y).

\displaystyle OH \perp PR \Rightarrow m_{OH} = m_{GQ} = \frac{18}{t}

\displaystyle m_{OH} = \frac{y-0}{14-0} = \frac{18}{t} \Rightarrow y=\frac{252}{t}

\displaystyle \therefore H\left(14, \frac{252}{t}\right)

(b) S lies on OP such that QS \perp OP and \angle PQS = \angle POQ.

(i) then

\displaystyle \frac{PS}{QS}=\frac{QS}{OS} \Rightarrow \frac{18}{t}=\frac{t}{32} \Rightarrow t^2=2\cdot 9 \cdot 2^5 \Rightarrow t=2^3\cdot 3 = 24

(ii) Now,

\displaystyle G\left(25, \frac{24^2-126}{24}\right) = \left(25,\frac{75}{4}\right) = (25, 18.75)

Q(32, 24)

\displaystyle m_{OQ} = \frac{24}{32}=\frac{3}{4}, \quad m_{OG} = \frac{75/4}{25} = \frac34, \quad m_{GQ} = \frac{18}{t} = \frac{18}{24} = \frac34

Therefore, OQ // OG // GQ and hence O, G, and Q are collinear.

(iii)

OP = 50

OR = \sqrt{14^2+48^2} = 2\sqrt{7^2+24^2} = 50

hence, △OPR is isosceles such that OP = OR.

Since Q is midpoint of PR, by SSS,

\triangle OQR \cong \triangle OQP

hence, H and I lie on OQ.

Since RQ = PQ and they are the altitudes of △GHR and △IPQ respectively, then

\displaystyle \frac{\text{Area of }\triangle GHR}{\text{Area of } \triangle IPQ} = \frac{HG}{IQ}

\displaystyle HG = \sqrt{(25-14)^2 + (18.75-10.5)^2} = \sqrt{11^2 + 8.25^2} = \sqrt{11^2+\left(\frac{33}{4}\right)^2}

\displaystyle = \frac{\sqrt{11^2 \cdot 16 + 33^2}}{4} = \frac{11}{4}\sqrt{16+9} = \frac{55}{4}

b = PQ = \sqrt{(50-32)^2+24^2} = \sqrt{18^2+24^2} = 6\sqrt{3^2+4^2} = 30

a = 50-b=50-30=20

So, I(20, y),

\displaystyle m_{OI} = m_{OQ} = \frac34 \Rightarrow \frac{y}{20}=\frac34 \Rightarrow y=15

\displaystyle IQ=\sqrt{(32-20)^2+(24-15)^2} = \sqrt{12^2+9^2} = 3\sqrt{4^2+3^2} = 15

\displaystyle \frac{HG}{IQ} = \frac{55/4}{15} = \frac{11}{12}

Therefore the ratio of the area of △GHR to the area of △IPQ is 11:12.