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In right triangle $ABC$ ($\angle A=90$), $AD$ is a height and $E$ is a point on $AC$ so that $BE=EC$ and $CD=CE=EB=1$. $\color {red} {Without}$ using trigonometric relations find $AE$.

enter image description here I do have a solution USING trigonometric relations ( $AE=\sqrt[3]{2}-1$ ),but it seems other solutions are troublesome.My attempt led to a complicated polynomial equation...

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    Is it ok to use the Pythagorean Theorem? (considering whether you consider this a "trigonometric relation" ...) – 2017-02-10
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    Yes of course,by trigonometric relations I mean $sin,cos$... – 2017-02-10

2 Answers 2

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Let $AE=x$.

Hence, $AD=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x}$ and since $AD^2=BD\cdot DC$,

we obtain $BD=x^2+2x$, which says that $AB=\sqrt{(x^2+2x)^2+x^2+2x}$ and $$BE=\sqrt{(x^2+2x)^2+x^2+2x+x^2}.$$ Thus,$$(x^2+2x)^2+x^2+2x+x^2=1$$ or $$x^4+4x^3+6x^2+2x-1=0$$ or $$(x+1)(x^3+3x^2+3x-1)=0$$ or $$(x+1)^3=2$$ or $$x=\sqrt[3]2-1$$

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    Bravo,you are brave enough to attack higher order equations:) – 2017-02-10
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    @Hamid Reza Ebrahimi Just the root $-1$ we see immediately. – 2017-02-10
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    "Hence, $DC$" should be "Hence, $AD$". You can dispense with the square roots by writing $1=BE^2=AB^2+x^2=AD^2+BD^2+x^2=AD^2+AD^4+x^2$, then plug $AD^2=x^2+2x$. – 2017-02-10
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    @grand_chat I fixed my post. Thank you! – 2017-02-10
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Here is a proof with only Pythagoreans:

(as I see, Michael Rozenberg used also the altitude theorem of right triangles)

We have 5 eqns in 5 variables AE, AC, BA, BD, AD:

Obviously $AE = AC - CE = AC - 1;$

and 4 Pythagoreans:

$$ AC^2 = - AB^2 + (BD+DC)^2 = - AB^2 + (BD+1)^2;\\ AC^2 = AD^2 + DC^2 = AD^2 + 1;\\ AD^2 + BD^2 = AB^2;\\ 1 = BE^2 = AB^2 + AE^2; $$

Using the third one to eliminate AB everywhere:

$$ AE = AC - 1;\\ AC^2 = - AD^2 - BD^2 + (BD+1)^2;\\ AC^2 = AD^2 + 1;\\ AD^2 + BD^2 = 1 - AE^2; $$

Using the third one of this block to eliminate AD everywhere:

$$AE = AC - 1;\\ 2 AC^2 = 1 - BD^2 + (BD+1)^2;\\ AC^2 + BD^2 = 2 - AE^2; $$

Using the first one of this block to eliminate AC everywhere:

$$2 (AE +1)^2 = 1 - BD^2 + (BD+1)^2 = 2 + 2 BD;\\ BD^2 = 2 - AE^2 - (AE + 1)^2;$$

Using the second one of this block to eliminate $BD^2$:

$$((AE +1)^2 -1 )^2 = 2 - AE^2 - (AE + 1)^2;$$

or, with $AE = x-1$,

$$ 0 = -(x^2 -1 )^2 + 2 - (x-1)^2 - x^2 = x (2 - x^3)$$

So we have the only positive real solution $AE = \sqrt[3] 2 -1$

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    How did you solve the last equation? – 2017-02-10
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    @HamidRezaEbrahimi I put it in the answer. – 2017-02-10
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    What if you did NOT know the answer before?? – 2017-02-10
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    Well, same business as the other answer. We both arrive at a fourth order equation, be it for AE or (AE +1). The only way to go about that (if you don't have computers available) is to guess one solution and split off the relevant factor. The form I got, $((AE +1)^2 -1 )^2 = 2 - AE^2 - (AE + 1)^2$, suggest to try and set $x = AE + 1$. Again, in any case, you have to guess one solution and $AE=-1$ is the easiest to see. – 2017-02-10
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    Thank you for your contribution:) (+1) – 2017-02-10