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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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    Easily? Looks to me like it's going to be a mess, though I confess I haven't actually tried it.2012-08-16
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    Anyway the equation is atmost quartic, and there are formulas to solve that and thats what i meant by 'easily'2012-08-16
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    Yes, it's quartic. I just expanded this out, and factored it by hand. It has two positive roots, both integers, and two irrational negative roots. Three of the four roots don't fit the figure geometrically. So there's just one solution to the geometric problem. As I said in my answer, the best way to find it is with some inspired guessing, based on well-known small Pythagorean triples.2012-08-16
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    @DavidWallace: How do you know the answer is an integer?2012-08-16
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    @user1729, you don't know it, but it can't hurt to try it.2012-08-17
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    @GerryMyerson: 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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    That's not maths!2012-08-16
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    @user1729: Actually sampling methods are indeed math, are used quite often, and even work in this instance. I'm not sure why anyone would discourage trying to get an intuitive feel for the problem.2012-08-16
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    Hey, @user1729, for any problem like this, it's worth TRYING a few things first, before getting into the hard algebra. In this case, it's comparatively easy to see what ONE solution is. Of course, proving that it's the ONLY solution is still hard - see my comment under pritam's answer. Getting a feel for any given problem is ABSOLUTELY part of maths.2012-08-17
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    @DavidWallace: Your comments under Pritram's answer boil down to "the way you prove the solution is unique is factorise the polynomial and show that all the other answers don't make sense". This finds all the solutions for you, and so your guessing was just adding to your work...2012-08-17
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    No. Knowing one of the solutions already made factorising the quartic much easier. Solving quartics is really hard - there are very many steps. But having seen one solution by using my intuition effectively turned it into a cubic; and solving cubics is substantially easier. So my "guessing" as you put it saved me lots of work.2012-08-17
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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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    right... And now?2012-08-16
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    I see... Now $x + 7 \gt 0$ magically becomes $x \gt 7$.2012-08-16
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    Now, we can look for triplets to approximate the triangle as close as possible to a near by triplet.2012-08-16
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    Here, the closest triplet is 8,15,17 for $x = 11$2012-08-16
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    The other triplet is 6,8,10 for the $\triangle$ EBC.2012-08-16
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    Why must the answer be an integer?2012-08-16
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    i just looked for an integer to start with and i got the answer2012-08-16
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    Is it then obvious that this answer is unique?2012-08-16
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    You have reasoned that this is the unique integer solution. But...why isn't, say, $5+\pi$ a solution?2012-08-16
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    What does the figure being to scale have to do with anything?2012-08-16