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Find $x$ in the following figure. enter image description here $AB,AC,AD,BC,BE,CD$ are straight lines.

$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$

$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$

NOTE: figure not to scale.

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    Please edit the question into the body of your post.2012-08-16

5 Answers 5

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By the Pythagorean theorem we have

$\begin{equation*} CE=\sqrt{10^{2}-\left( x-3\right) ^{2}}=\sqrt{91-x^{2}+6x} \end{equation*}$ and $\begin{equation*} CD^{2}+AD^{2}=AC^{2}=\left( CE+AE\right) ^{2} \end{equation*}.$ So we have to solve the following irrational equation $\begin{equation*} \left( x-3\right) ^{2}+\left( x+4\right) ^{2}=\left( \sqrt{91-x^{2}+6x} +x\right) ^{2},\tag{1} \end{equation*}$ which can be simplified to $\begin{equation*} x^{2}-2x-33=\sqrt{-x^{4}+6x^{3}+91x^{2}}. \end{equation*}$

After squaring both sides and grouping the terms of the same degree we get the quartic equation $\begin{equation*} 2x^{4}-10x^{3}-153x^{2}+132x+1089=0.\tag{2} \end{equation*}$

The coefficient of $x^{4}$ is $2=1\times 2$ and the constant term is $1089=1\times 3^{2}11^{2}$. To find possible rational roots of this equation, we apply the rational root theorem and test the numbers of the form $\begin{equation*} x=\pm \frac{p}{q}, \end{equation*}$ where $p\in \left\{ 1,3,9,11,33,99,121,363,1089\right\} $ is a divisor of $1089$ and $q\in \left\{ 1,2\right\} $ is a divisor of $2$. It turns out that $x=3$ and $x=11$ are roots. Now we divide the LHS by $x-3$ $ \begin{equation*} \frac{2x^{4}-10x^{3}-153x^{2}+132x+1089}{x-3}=2x^{3}-4x^{2}-165x-363 \end{equation*}$ and this quotient by $x-11$ $\begin{equation*} \frac{2x^{3}-4x^{2}-165x-363}{x-11}=2x^{2}+18x+33. \end{equation*}$ So we have the equivalent equation $\begin{equation*} \left( x-3\right) (x-11)\left( 2x^{2}+18x+33\right) =0\tag{3} \end{equation*}$ Since the solutions of $2x^{2}+18x+33$ are both negative and $x=3$ is not a solution of the original irrational equation, the solution is therefore $x=11. $

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Hint: Using Pythagorean theorem $(x+4)^2+(x-3)^2=\left( x+\sqrt{10^2-(x-3)^2}\right)^2$ and this can be easily solved.

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    @$G$erryMyerson: Sure, it can't hurt. You might find an answer. But essentially guessing won't tell you if the answer is unique. Using pritam's method gives you uniqueness. Anyway, if guessing is allowed, then so is using [wolfram-alpha](http://www.wolframalpha.com/) to factorise this! You've done the maths, so let a computer do the plug-and-chug bit.2012-08-17
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It just so happens that one solution is an integer. So maybe, you could try a few numbers, and see if any of them jump out as the solution, before you set about trying to solve a nasty quartic. Focus on well-known small Pythagorean triples.

Note that it took me less than a minute of staring at the figure, to realise what the solution was. I don't yet know whether there are any other solutions that fit the figure.

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    @DavidWallace: That is true, and it does make it easier. However, it is not what you said in your comment!2012-08-19
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$∆BCE$ is right angle triangle.

Hence $BC^2 = BE^2 + EC^2$ $EC = \sqrt{(BC^2 - BE^2})= \sqrt{(100 - (x-3)^2)}$

$∆ACD$ is right angle triangle.

Hence $AC^2 = CD^2 + AD^2$

$(AE + EC)^2 = CD^2 + AD^2$

Substitute the values, $(\sqrt{(100 - (x-3)^2)} + x)^2 = (x-3)^2 + (x+4)^2$

Then you can solve this equation easily for getting x value.

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"The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides".

By the Triangle Inequality Theorem,

$CE + (x-3) \gt 10$, $CE + x \lt (x+4) + (x-3)$

i.e. $ (13-x) \lt CE \lt (x+1)$

we now have, $(13-x) \lt (x+1)$

i.e. $x \gt 6$

from the, $\triangle EBC $ we have,

$x-3 \lt 10$

i.e. $x \lt 13$

we can conclude that, $6 \lt x \lt 13$

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    What does the figure being to scale have to do with an$y$thing?2012-08-16