Find the value of $x$ in the displayed figure

Question

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.

Answer 1

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 $$x=11. $$

Answer 2

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.

Answer 3

If you were given that $x$ and |EC| were integers you could use the following.

Let $y = |EC|$.

Using Pythagoras on $\triangle BEC$:

1) $(x-3)^2 + y^2 = 10^2$ $\Rightarrow x^2 - 6x + 9 + y^2 = 100$ $\Rightarrow y^2 = -x^2 + 6x + 91$

Using Pythagorus on $\triangle ACD$:

2) $(x+4)^2 + (x-3)^2 = (x+y)^2$ $\Rightarrow x^2 + 2x + 25 = 2xy + y^2$

Then put 1) into 2):

$\Rightarrow x^2 + 2x + 25 = 2xy + (-x^2 + 6x +91)$ $ \Rightarrow xy = x^2 - 2x - 33$

Then: $y = x - 2 - \frac{33}{x}$

So $x$ has to be a divisor of 33: 11, 3 or 1.

I tried using the cosine rule then but just ended up with 0 = 0. The trickiest part is the angle $ABC$ just about failing to be a right angle.

Answer 4

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.

Answer 5

$∆BCE$ 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.

Answer 6

"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, $6 \lt x \lt 13$