Find $x$ in the following figure.
$AB,AC,AD,BC,BE,CD$ 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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asked Aug 16, 2012 at 5:14
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$\begingroup$ Please edit the question into the body of your post. $\endgroup$– Gerry MyersonAug 16, 2012 at 5:58
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5 Answers
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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 $$x=11. $$
answered Aug 16, 2012 at 14:52
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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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3$\begingroup$ Easily? Looks to me like it's going to be a mess, though I confess I haven't actually tried it. $\endgroup$ Aug 16, 2012 at 6:00
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$\begingroup$ Anyway the equation is atmost quartic, and there are formulas to solve that and thats what i meant by 'easily' $\endgroup$– pritamAug 16, 2012 at 6:04
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$\begingroup$ 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. $\endgroup$– user22805Aug 16, 2012 at 7:38
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1$\begingroup$ @DavidWallace: How do you know the answer is an integer? $\endgroup$– user1729Aug 16, 2012 at 12:37
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$\begingroup$ @user1729, you don't know it, but it can't hurt to try it. $\endgroup$ Aug 17, 2012 at 6:19
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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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2$\begingroup$ @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. $\endgroup$– ex0du5Aug 16, 2012 at 20:32
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$\begingroup$ 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. $\endgroup$– user22805Aug 17, 2012 at 3:08
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$\begingroup$ @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... $\endgroup$– user1729Aug 17, 2012 at 9:28
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$\begingroup$ 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. $\endgroup$– user22805Aug 17, 2012 at 9:33
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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, $6 \lt x \lt 13$
answered Aug 16, 2012 at 7:47
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$\begingroup$ I see... Now $x + 7 \gt 0$ magically becomes $x \gt 7$. $\endgroup$– t.b.Aug 16, 2012 at 8:19
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$\begingroup$ Now, we can look for triplets to approximate the triangle as close as possible to a near by triplet. $\endgroup$ Aug 16, 2012 at 8:24
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2$\begingroup$ You have reasoned that this is the unique integer solution. But...why isn't, say, $5+\pi$ a solution? $\endgroup$– user1729Aug 16, 2012 at 13:06
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4$\begingroup$ What does the figure being to scale have to do with anything? $\endgroup$– EmilyAug 16, 2012 at 15:54