Tek at 13:34, 8 October 2008

2008-10-08T13:34:12Z

Ian at 10:14, 28 September 2008

2008-09-28T10:14:49Z

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		+
		+
		+
		+
		+	First, let's decide to determine all distance in dm (decimeters), so that we have all the distances as integers.
		+
		+	Call the length of the washing line from the trees to the hanger
		+	<math>y</math>
		+	and
		+	<math>z</math>, as in the figure below, and introduce two auxiliary triangles which have
		+	<math>y</math>
		+	and
		+	<math>z</math>
		+	as their hypotenuses. (As an approximation, we suppose that the taut washing line consists of two straight parts.)
		+
		+	Because the line is
		+	<math>\text{54}</math>
		+	dm long, we have
		+
		+
		+	<math>y+z=54</math>
		+	(1)
		+
		+	Then, Pythagoras' theorem gives the relations
		+
		+
		+	<math>y^{2}=x^{2}+12^{2}</math>
		+	(2)
		+
		+	<math>z^{2}=\left( x+6 \right)^{2}+36^{2}</math>
		+	(3)
		+
		+
		+	The idea now is to solve the equation system (1)-(3) by first eliminating
		+	<math>z</math>, so that we get two equations which only contain
		+	<math>x</math>
		+	and
		+	<math>y</math>. Then, eliminate
		+	<math>y</math>
		+	from one of these equations, so that we get an equation which determines
		+	<math>x</math>.
		+
		+	From (1), we have
		+	<math>z=\text{54}-y</math>, and substituting this into (3) gives us the equation
		+
		+
		+	<math>\left( \text{54}-y \right)^{2}=\left( x+6 \right)^{2}+36^{2}</math>
		+	(3')
		+
		+	Equations (2) and (3') together give a smaller equation system for
		+	<math>x</math>
		+	and
		+	<math>y</math>:
		+
		+
		+
		+	<math>\left\{ \begin{align}
		+	& y^{2}=x^{2}+12^{2}\quad \quad \quad \quad \quad \quad \quad (2) \\
		+	& \left( \text{54}-y \right)^{2}=\left( x+6 \right)^{2}+36^{2}\quad \quad \left( \text{3}' \right) \\
		+	\end{align} \right.</math>
		+
		+
		+
		+
		+	Expand the quadratic terms on both sides of (3')
		+
		+
		+	<math>54^{2}-2\centerdot 54\centerdot y+y^{2}=x^{2}+2\centerdot 6\centerdot x+6^{2}+36^{2}</math>
		+
		+
		+	and simplify
		+
		+
		+	<math>2916-108y+y^{2}=x^{2}+12x+1332</math>
		+
		+
		+	Use (2) and replace
		+	<math>y^{\text{2}}</math>
		+	with
		+	<math>x^{\text{2}}+\text{12 }</math>
		+	in this equation,
		+
		+
		+	<math>2916-108y+x^{2}+144=x^{2}+12x+1332</math>
		+
		+
		+	which gets rid of the
		+	<math>x^{\text{2}}\text{ }</math>
		+	term ,
		+
		+
		+	<math>2916-108y+144=12x+1332</math>
		+
		+
		+	and further simplification gives the equation
		+
		+
		+	<math>12x+108y=1728\quad \quad \quad \quad \quad \left( \text{3}'' \right)</math>
		+
		+
		+	If we pause for a moment and summarize the situation, we see that we have succeeded in simplifying the equation system (2) and (3') to a system (2) and
		+	<math>\left( \text{3}'' \right)</math>, where one of the equations is linear
		+
		+
		+	<math>\left\{ \begin{align}
		+	& y^{2}=x^{2}+12^{2}\quad \quad \quad \quad \quad \quad \quad (2) \\
		+	& 12x+108y=1728\quad \quad \quad \quad \quad \left( \text{3}'' \right) \\
		+	\end{align} \right.</math>
		+
		+
		+
		+	In this system, we can make
		+	<math>y\text{ }</math>
		+	the subject in
		+	<math>\left( \text{3}'' \right)</math>,
		+
		+
		+	<math>y=\frac{1728-12x}{108}=16-\frac{x}{9}</math>
		+
		+
		+	and substitute into (2):
		+
		+
		+	<math>\left( 16-\frac{x}{9} \right)^{2}=x^{2}+144</math>
		+
		+
		+	This is an equation which only contains
		+	<math>x</math>, and if we solve it, we will get our answer.
		+
		+	Expand the quadratic on the left-hand side,
		+
		+
		+	<math>16^{2}-2\centerdot 16\centerdot \frac{x}{9}+\left( \frac{x}{9} \right)^{2}=x^{2}+144</math>
		+
		+
		+
		+	and collect together all terms on one side,
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-\frac{x^{2}}{81}+\frac{32}{9}x+144-16^{2}=0 \\
		+	& \\
		+	\end{align}</math>
		+
		+	which gives the equation
		+
		+
		+	<math>\frac{80}{81}x^{2}+\frac{32}{9}x-112=0</math>
		+
		+
		+	Multiply both sides by
		+	<math>\frac{81}{80}</math>
		+	so that we get the equation in standard form,
		+
		+
		+	<math>x^{2}+\frac{18}{5}x-\frac{567}{5}=0</math>
		+
		+
		+	Completing the square on the left-hand side gives
		+
		+
		+	<math>\left( x+\frac{9}{5} \right)^{2}-\left( \frac{9}{5} \right)^{2}-\frac{567}{5}=0</math>
		+
		+	and then
		+
		+
		+	<math>\left( x+\frac{9}{5} \right)^{2}=\frac{81}{25}+\frac{567}{5}=\frac{2916}{25}</math>
		+
		+
		+	i.e.
		+
		+
		+	<math>x=-\frac{9}{5}\pm \sqrt{\frac{2916}{25}}=-\frac{9}{5}\pm \frac{54}{5}</math>
		+
		+
		+	This means that the equation has the solutions
		+
		+
		+	<math>x=-\frac{9}{5}-\frac{54}{5}=-\frac{63}{5}</math>
		+	and
		+
		+	<math>x=-\frac{9}{5}+\frac{54}{5}=9</math>
		+
		+
		+	The answer is thus
		+	<math>x=\text{9 }</math>
		+	dm (the negative root must be discarded).
		+
		+	To be sure that we have calculated correctly, we also look at the values of
		+	<math>y</math>
		+	and
		+	<math>z</math>, and check that the original equations (1) to (3) are satisfied.
		+
		+	Equation (3'') gives
		+
		+
		+	<math>y=16-\frac{x}{9}=16-1=15</math>
		+
		+
		+	and equation (1) gives
		+
		+
		+	<math>z=54-y=54-15=39</math>
		+
		+
		+	Now, we check that x=9,
		+	<math>y</math>
		+	=15 and z=39 satisfy the equations (1),(12) and (3).
		+
		+	LHS of (1)
		+	<math>=15+39=54</math>
		+
		+	RHS of (1)
		+	<math>=54</math>
		+
		+	LHS of (2)
		+	<math>=15^{2}=225</math>
		+
		+	RHS of (2)
		+	<math>=9^{2}+12^{2}=81+144=225</math>
		+
		+	LHS of (3)
		+	<math>=39^{2}=1521</math>
		+
		+	RHS of (3)
		+	<math>=\left( 9+6 \right)^{2}+36^{2}=15^{2}+36^{2}=225+1296=1521</math>

Tekbot: Lösning 4.1:10 moved to Solution 4.1:10: Robot: moved page

2008-09-09T09:33:23Z

Lösning 4.1:10 moved to Solution 4.1:10: Robot: moved page

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Revision as of 09:33, 9 September 2008

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2008-08-21T07:03:49Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Martin at 14:44, 15 August 2008

2008-08-15T14:44:15Z

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2008-04-02T13:35:20Z

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