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# Some algebra questions

+1
513
20
+7076

1) A frog started from the origin of the coordinate plane and made three jumps. Each time the frog jumped a distance of 5 units and landed at a point with integer coordinates. How many different possibilities of the final position of the frog are there?

2) Find the product of the real roots to the equation $$x^2+7x-5=5\sqrt{x^3-1}$$.

3) A real number $$a$$ is randomly chosen in the interval$$-20\leq a\leq 18$$. What is the probability that all roots of the equation $$x^3+(2a+1)x^2+(4a-1)x+2=0$$ are real?

4) Expanding $$(x^2+3x+2)^2$$ and simplifying (collecting like terms) gives $$x^4+6x^3+13x^2+12x+4$$, which consists of 5 terms. If we expand $$(a^2+20ab+18)^{2018}$$ and simplify, how many terms will there be?

Source: International Mathematical Olympiad Preliminary Selection Contest - Hong Kong 2018.

I was in the contest, and I struggled in these problems. There are 20 questions, and only 3 hours to do them all, so it is around 9 minutes per question. It is so hard! :(

(P.S. I don't know the answers.)

May 24, 2018
edited by MaxWong  May 24, 2018

#1
+2

Hi Max. Sorry, no LaTex!

2.         Solve for x:

x^2 + 7 x - 5 = 5 sqrt(x^3 - 1)

x^2 + 7 x - 5 = 5 sqrt(x^3 - 1) is equivalent to 5 sqrt(x^3 - 1) = x^2 + 7 x - 5:

5 sqrt(x^3 - 1) = x^2 + 7 x - 5

Raise both sides to the power of two:

25 (x^3 - 1) = (x^2 + 7 x - 5)^2

Expand out terms of the left hand side:

25 x^3 - 25 = (x^2 + 7 x - 5)^2

Expand out terms of the right hand side:

25 x^3 - 25 = x^4 + 14 x^3 + 39 x^2 - 70 x + 25

Subtract x^4 + 14 x^3 + 39 x^2 - 70 x + 25 from both sides:

-x^4 + 11 x^3 - 39 x^2 + 70 x - 50 = 0

The left hand side factors into a product with three terms:

-(x^2 - 8 x + 10) (x^2 - 3 x + 5) = 0

Multiply both sides by -1:

(x^2 - 8 x + 10) (x^2 - 3 x + 5) = 0

Split into two equations:

x^2 - 8 x + 10 = 0 or x^2 - 3 x + 5 = 0

Subtract 10 from both sides:

x^2 - 8 x = -10 or x^2 - 3 x + 5 = 0

x^2 - 8 x + 16 = 6 or x^2 - 3 x + 5 = 0

Write the left hand side as a square:

(x - 4)^2 = 6 or x^2 - 3 x + 5 = 0

Take the square root of both sides:

x - 4 = sqrt(6) or x - 4 = -sqrt(6) or x^2 - 3 x + 5 = 0

x = 4 + sqrt(6) or x - 4 = -sqrt(6) or x^2 - 3 x + 5 = 0

x = 4 + sqrt(6) or x = 4 - sqrt(6) or x^2 - 3 x + 5 = 0

Subtract 5 from both sides:

x = 4 + sqrt(6) or x = 4 - sqrt(6) or x^2 - 3 x = -5

x = 4 + sqrt(6) or x = 4 - sqrt(6) or x^2 - 3 x + 9/4 = -11/4

Write the left hand side as a square:

x = 4 + sqrt(6) or x = 4 - sqrt(6) or (x - 3/2)^2 = -11/4

Take the square root of both sides:

x = 4 + sqrt(6) or x = 4 - sqrt(6) or x - 3/2 = (i sqrt(11))/2 or x - 3/2 = 1/2 (-i) sqrt(11)

x = 4 + sqrt(6) or x = 4 - sqrt(6) or x = 3/2 + (i sqrt(11))/2 or x - 3/2 = -(i sqrt(11))/2

x = 4 + sqrt(6) or x = 4 - sqrt(6) or x = 3/2 + (i sqrt(11))/2 or x = 3/2 - (i sqrt(11))/2

[4 + sqrt(6)] * [4 - sqrt(6)]= 10

[Courtesy of Mathematica 11 Home Edition]

May 24, 2018
#2
+2

Welcome back max ;)

Here's a hint for 3: -2 will always be a root of the polynomial

EDIT: whoops, tried to post this as an answer

Guest May 24, 2018
edited by Guest  May 24, 2018
#3
+9797
+2

12^3=1728

May 24, 2018
#4
+2

They're not asking for the number of possible paths the frog can take, they're asking for the number of possible final positions

Guest May 24, 2018
#5
+2

This is what I think!

3! x 4 = 24 final positions - not including its rest position at "0"!

May 24, 2018
#16
0

This is pure moronic BULLSHlT!!!!!!!!!!

Guest May 27, 2018
edited by Guest  May 27, 2018
#9
+20847
+1

1)
A frog started from the origin of the coordinate plane and made three jumps.
Each time the frog jumped a distance of 5 units and landed at a point with integer coordinates.
How many different possibilities of the final position of the frog are there?

Startposition ( 0, 0 ):
After 1. Jump, there are 12 different positions with a distance of 5 units.
1.) ( 5, 0 )
2.) ( 4, 3 )
3.) ( 3, 4 )
4.) ( 0, 5 )
5.) ( -3, 4 )
6.) ( -4, 3 )
7.) ( -5, 0 )
8.) ( -4, -3 )
9.) ( -3, -4 )
10.) ( 0, -5 )
11.) ( 3, -4 )
12.) ( 4, -3 )

There are $$12^3$$ = 1728  jumps. But 264 different possibilities of the final position.

1.) (15, 0)

2.) (14, 3)

3.) (13, 4)

4.) (10, 5)

5.) (7, 4)

6.) (6, 3)

7.) (5, 0)

8.) (6, -3)

9.) (7, -4)

10.) (10, -5)

11.) (13, -4)

12.) (14, -3)

13.) (13, 6)

14.) (12, 7)

15.) (9, 8)

16.) (6, 7)

17.) (5, 6)

18.) (4, 3)

19.) (6, -1)

20.) (9, -2)

21.) (12, -1)

22.) (13, 0)

23.) (11, 8)

24.) (8, 9)

25.) (5, 8)

26.) (4, 7)

27.) (3, 4)

28.) (4, 1)

29.) (8, -1)

30.) (11, 0)

31.) (12, 1)

32.) (5, 10)

33.) (2, 9)

34.) (1, 8)

35.) (0, 5)

36.) (1, 2)

37.) (2, 1)

38.) (8, 1)

39.) (9, 2)

40.) (-1, 8)

41.) (-2, 7)

42.) (-3, 4)

43.) (-2, 1)

44.) (-1, 0)

45.) (2, -1)

46.) (6, 1)

47.) (-3, 6)

48.) (-4, 3)

49.) (-3, 0)

50.) (-2, -1)

51.) (1, -2)

52.) (4, -1)

53.) (-5, 0)

54.) (-4, -3)

55.) (-3, -4)

56.) (0, -5)

57.) (3, -4)

58.) (4, -3)

59.) (-3, -6)

60.) (-2, -7)

61.) (1, -8)

62.) (4, -7)

63.) (5, -6)

64.) (-1, -8)

65.) (2, -9)

66.) (5, -8)

67.) (6, -7)

68.) (5, -10)

69.) (8, -9)

70.) (9, -8)

71.) (11, -8)

72.) (12, -7)

73.) (13, -6)

74.) (12, 9)

75.) (11, 10)

76.) (8, 11)

77.) (4, 9)

78.) (3, 6)

79.) (5, 2)

80.) (11, 2)

81.) (12, 3)

82.) (10, 11)

83.) (7, 12)

84.) (4, 11)

85.) (3, 10)

86.) (2, 7)

87.) (7, 2)

88.) (10, 3)

89.) (11, 4)

90.) (4, 13)

91.) (1, 12)

92.) (0, 11)

93.) (1, 4)

94.) (8, 5)

95.) (-2, 11)

96.) (-3, 10)

97.) (-4, 7)

98.) (-2, 3)

99.) (5, 4)

100.) (-4, 9)

101.) (-5, 6)

102.) (-3, 2)

103.) (0, 1)

104.) (3, 2)

105.) (-6, 3)

106.) (-4, -1)

107.) (-1, -2)

108.) (3, 0)

109.) (-2, -5)

110.) (1, -6)

111.) (4, -5)

112.) (5, -4)

113.) (7, -6)

114.) (8, -5)

115.) (11, -4)

116.) (12, -3)

117.) (9, 12)

118.) (6, 13)

119.) (3, 12)

120.) (2, 11)

121.) (2, 5)

122.) (9, 4)

123.) (3, 14)

124.) (0, 13)

125.) (-1, 12)

126.) (-2, 9)

127.) (-1, 6)

128.) (6, 5)

129.) (7, 6)

130.) (-3, 12)

131.) (-4, 11)

132.) (-5, 8)

133.) (-4, 5)

134.) (0, 3)

135.) (4, 5)

136.) (-5, 10)

137.) (-6, 7)

138.) (-5, 4)

139.) (-1, 2)

140.) (2, 3)

141.) (-7, 4)

142.) (-6, 1)

143.) (1, 0)

144.) (-5, -2)

145.) (-1, -4)

146.) (2, -3)

147.) (3, -2)

148.) (3, -6)

149.) (6, -5)

150.) (9, -4)

151.) (10, -3)

152.) (11, -2)

153.) (0, 15)

154.) (-3, 14)

155.) (-4, 13)

156.) (-6, 13)

157.) (-7, 12)

158.) (-8, 9)

159.) (-7, 6)

160.) (-6, 5)

161.) (1, 6)

162.) (-8, 11)

163.) (-9, 8)

164.) (-8, 5)

165.) (-1, 4)

166.) (-10, 5)

167.) (-9, 2)

168.) (-8, 1)

169.) (-8, -1)

170.) (-7, -2)

171.) (0, -1)

172.) (-6, -3)

173.) (0, -3)

174.) (7, -2)

175.) (-9, 12)

176.) (-10, 11)

177.) (-11, 8)

178.) (-9, 4)

179.) (-2, 5)

180.) (-11, 10)

181.) (-12, 7)

182.) (-11, 4)

183.) (-10, 3)

184.) (-7, 2)

185.) (-13, 4)

186.) (-12, 1)

187.) (-11, 0)

188.) (-4, 1)

189.) (-11, -2)

190.) (-10, -3)

191.) (-7, -4)

192.) (-3, -2)

193.) (-9, -4)

194.) (-6, -5)

195.) (-2, -3)

196.) (1, -4)

197.) (5, -2)

198.) (-12, 9)

199.) (-13, 6)

200.) (-12, 3)

201.) (-11, 2)

202.) (-5, 2)

203.) (-14, 3)

204.) (-13, 0)

205.) (-12, -1)

206.) (-9, -2)

207.) (-6, -1)

208.) (-12, -3)

209.) (-11, -4)

210.) (-8, -5)

211.) (-5, -4)

212.) (-10, -5)

213.) (-7, -6)

214.) (-4, -5)

215.) (-4, -7)

216.) (-1, -6)

217.) (2, -5)

218.) (-15, 0)

219.) (-14, -3)

220.) (-13, -4)

221.) (-13, -6)

222.) (-12, -7)

223.) (-9, -8)

224.) (-6, -7)

225.) (-5, -6)

226.) (-11, -8)

227.) (-8, -9)

228.) (-5, -8)

229.) (-5, -10)

230.) (-2, -9)

231.) (2, -7)

232.) (-12, -9)

233.) (-11, -10)

234.) (-8, -11)

235.) (-4, -9)

236.) (-10, -11)

237.) (-7, -12)

238.) (-4, -11)

239.) (-3, -10)

240.) (-4, -13)

241.) (-1, -12)

242.) (0, -11)

243.) (2, -11)

244.) (3, -10)

245.) (4, -9)

246.) (-9, -12)

247.) (-6, -13)

248.) (-3, -12)

249.) (-2, -11)

250.) (-3, -14)

251.) (0, -13)

252.) (1, -12)

253.) (3, -12)

254.) (4, -11)

255.) (0, -15)

256.) (3, -14)

257.) (4, -13)

258.) (6, -13)

259.) (7, -12)

260.) (8, -11)

261.) (9, -12)

262.) (10, -11)

263.) (11, -10)

264.) (12, -9)

May 25, 2018
edited by heureka  May 25, 2018
#11
0

Start position ( 0, 0 ):
After 1. Jump, there are 12 different positions with a distance of 5 units

12C2 = 66 x 4 = 264 final positions.

Guest May 25, 2018
#12
+1

Guest May 25, 2018
edited by Guest  May 25, 2018
edited by Guest  May 25, 2018
#17
0

This is pure BULLSHlT!!!!!!!!!!

Need proof? Here are some combinations and variations as evidence.

PURE SHlT FROM A BULL, THIS IS

THIS IS PURE SHlT FROM A BULLSHITTER

FROM A BULLSHITTER, THIS IS PURE SHlT

Deleting this post doesn’t change that.

Guest May 27, 2018
#10
+1

4.

(a^2 + 2ab + 18)^2018

For all exponents n, the number of terms =1/2(n + 2 - 1)(n + 3 - 1). Example: 5th power:

1/2(5 + 2 -1)(5 + 3 - 1) = 1/2 x 6 x 7 = 21 terms. Therefore, to the power of 2018:

1/2 (2018 + 2 - 1)(2018 + 3 - 1) = 1/2 x 2019 x 2020 =2,039,190 number of terms.

May 25, 2018
#13
+1

3.

x^3+(2a+1)x^2+(4a-1)x+2=0

x = -2, ALL values of "a" are solutions.

x = - 1/2,  a = 7/4

x = 1/2,    a = - 3/4

Since ALL the solutions are "real", the probability is 100%!!.

May 25, 2018
#14
+7076
+1

I think the question is asking all roots(x) are real, not a.

Therefore, shouldn't we solve the general root first, then find the probability that all x are real?

I mean:

$$x^3+(2a+1)x^2+(4a-1)x+2 = 0\\ x = -2 , \dfrac{\sqrt{4a^2-4a-3}-2a+1}{2}, \dfrac{-\sqrt{4a^2-4a-3}-2a+1}{2}$$

So the question becomes probability that $$\sqrt{4a^2-4a-3}$$ is real.

MaxWong  May 26, 2018
#18
+20847
+2

1)
A frog started from the origin of the coordinate plane and made three jumps.
Each time the frog jumped a distance of 5 units and landed at a point with integer coordinates.
How many different possibilities of the final position of the frog are there?

All 264 different possibilities of the final position of the frog:

\scalebox{0.5}{% scale
\begin{tikzpicture}
\draw[thick] (-15,-15) grid (15,15);
% Axes
\foreach \x in {-15,...,-1,0,1,2,...,15} {%
\draw (\x, -.1) -- (\x, 0) node[below left=2pt] {$\scriptstyle\x$};
}
\foreach \y in {-15,...,-1,1,2,...,15} {%
\draw (-.1,\y) -- (0,\y) node[below left=2pt] {$\scriptstyle\y$};
};

\foreach \x/\y in {5/0,4/3,3/4,0/5,-5/0,-4/3,-3/4,4/-3,3/-4,0/-5/,-4/-3/,-3/-4} %{\x \y,}
{
\foreach \v/\w in {5/0,4/3,3/4,0/5,-5/0,-4/3,-3/4,4/-3,3/-4,0/-5/,-4/-3/,-3/-4} %{\v \w,}
{
\foreach \r/\s in {5/0,4/3,3/4,0/5,-5/0,-4/3,-3/4,4/-3,3/-4,0/-5/,-4/-3/,-3/-4} %{\r \s,}
{
\node[draw,circle,inner sep=2pt,fill] at (\x+\v+\r,\y+\w+\s) {};
}
}
}
\end{tikzpicture}
}

May 28, 2018
edited by heureka  May 28, 2018
#19
+95347
+1

Very impressive Heureka

Melody  May 28, 2018
#20
+20847
+1

Thank you Melody!

heureka  May 28, 2018