computation, right triangle inside right triangle 2000 Portugal p5

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#1 h Oct 12, 2020, 5:53 PM

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In the figure, $[ABC]$ and $[DEC]$ are right triangles . Knowing that $EB = 1/2, EC = 1$ and $AD = 1$ , calculate $DC$ .
//cdn.artofproblemsolving.com/images/5/7/d/57d9fcf9e54cf73e094ef5a9853c894337706e7c.png

//cdn.artofproblemsolving.com/images/5/7/d/57d9fcf9e54cf73e094ef5a9853c894337706e7c.png

This post has been edited 2 times. Last edited by parmenides51, Oct 13, 2020, 2:11 AM

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Desimathematics

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Desimathematics

#3 h Oct 13, 2020, 3:30 PM

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deleted.....

This post has been edited 1 time. Last edited by Desimathematics, Oct 13, 2020, 6:32 PM

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#4 h Oct 13, 2020, 5:26 PM • 1 Y

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That's not quite right @above, $DE=\tan\theta$ and $DC=\sec\theta$ (as opposed to what you said)

$\cos\theta\sin (60-\theta)=\sin\theta\sin 30$
$\sin 60-\sin\theta\cos (60-\theta)=\sin\theta\sin 30$
$\sin 60=\sin\theta(\cos (60-\theta)+\sin 30)=\sin\theta(\cos 60\cos\theta+\sin 60\sin\theta+\sin 30)$
$\frac{\sqrt{3}}{2}=\frac{1}{2}\sin\theta\cos\theta+\frac{\sqrt{3}}{2}\sin^2\theta+\frac{1}{2}\sin\theta$
$\frac{\sqrt{3}}{2}\cos^2\theta=\frac{1}{2}\sin\theta(1+\cos\theta)$
$\frac{3\cos^4\theta}{1+2\cos\theta+\cos^2\theta}=1-\cos^2\theta$ . If $x=\cos\theta$
$3x^4=-x^4-2x^3+2x+1$
$f(x)=4x^4+2x^3-2x-1=0$
$f(-1)=3$
$f(1)=3$ , so there must be atleast 2 roots.

We see that $f(-.5)=0$ , however, that leads to an extranneous obtuse angle
$g(x)=\frac{f(x)}{2x+1}=2x^3-1$
So the other root is $\sqrt[3]{\frac{1}{2}}$
So $DC=\sec\theta=\sqrt[3]{2}$

This seems pretty hard to compute synthetically, as there is a cube root.

Anyway here is proof that my answer is correct: https://www.geogebra.org/calculator/kwbemrzt

This post has been edited 2 times. Last edited by natmath, Oct 13, 2020, 5:32 PM

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#5 h Oct 13, 2020, 6:27 PM • 1 Y

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Let $A(0,0),E(b,0),B(b+\frac{1}{2},0),C(b+\frac{1}{2},\frac{\sqrt{3}}{2})$ .

The line $ED\ :\ y=\frac{b-x}{\sqrt{3}}$ cuts the line $AC\ :\ y=\frac{\sqrt{3}}{2b+1}x$ in the point $D$ .
Point $D(\frac{b(2b+1)}{2b+4},\frac{b\sqrt{3}}{2b+4})$ .

Condition $AD=1$ gives $b^{4}+b^{3}-4b-4=0$ with solutions $b=-1,b=\sqrt[3]{4}$ .
Distance $AC=\sqrt[3]{2}+1$ , so $CD=\sqrt[3]{2}$ .

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