AB=AD+DC for cyclic ABCD, <ABC=60^o, BC = CD (2017 Polish JMO R1 p4)

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AB=AD+DC for cyclic ABCD, <ABC=60^o, BC = CD (2017 Polish JMO R1 p4)

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#1 h Apr 6, 2021, 5:36 PM • 1 Y

Y by samrocksnature

The quadrilateral $ABCD$ is inscribed in a circle where $\angle ABC = 60^o$ and $BC = CD$ . Prove $AB = AD + DC$ .

https://cdn.artofproblemsolving.com/attachments/4/a/467ebcd606cf26ab7c2ac3fe1c3e1aa55604e8.png

This post has been edited 1 time. Last edited by parmenides51, Apr 6, 2021, 5:37 PM

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#2 h Apr 6, 2021, 5:37 PM • 2 Y

Y by samrocksnature, Mango247

posted for the image link

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#3 h Apr 6, 2021, 5:47 PM • 1 Y

Y by samrocksnature

WRONG

We let $\angle CBD=\angle CDB = \theta$ .

We know that $\angle ADB=90^{\circ}$ so $90^{\circ}+60^{\circ}+\theta=180^{\circ}$ so $\theta=30^{\circ}$ .

We let $AD=x$ .

We know that $AB=2x$ and $BD=x\sqrt3$ so $DC=x$ .

Therefore, $2x=AB=AD+DC=x+x$ .

This post has been edited 1 time. Last edited by RedFireTruck, Apr 7, 2021, 1:59 AM

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BMOBoy

#4 h Apr 6, 2021, 6:48 PM

Y by

FireTruck, you assumed that AB was a diameter.

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#5 h Apr 6, 2021, 7:01 PM

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BMOBoy wrote:

FireTruck, you assumed that AB was a diameter.

oh yeah. oops...

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#6 h Apr 6, 2021, 7:02 PM

Y by

Note that arc $AC$ has measure $120$ .
If arc $DC$ has measure $x$ , then arc $AB$ has measure $240-x$ and arc $AD$ has measure $120-x$
The length of $AB$ is
$2r\sin (120-\frac{x}{2})$ $\boxed{2r\sin(60+\frac{x}{2})}$ The length of $AD+DC$ is
$2r(\sin(60-\frac{x}{2})+\sin\frac{x}{2})$ $2r(\sin(120+\frac{x}{2})+\sin\frac{x}{2})$ Using sum to product identities, this expression is equivalent to
$2r\sin(60+\frac{x}{2})$

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#7 h Apr 6, 2021, 7:13 PM

Y by

parmenides51 wrote:

The quadrilateral $ABCD$ is inscribed in a circle where $\angle ABC = 60^o$ and $BC = CD$ . Prove $AB = AD + DC$ .

https://cdn.artofproblemsolving.com/attachments/4/a/467ebcd606cf26ab7c2ac3fe1c3e1aa55604e8.png

Let $C'$ be a point on ray $AD$ such that $DC'=DC$ . Then $\angle CDC'=60$ and $\triangle CDC'$ is isosceles. Hence we can conclude that $\triangle CDC'$ is equilateral. Hence $CC'=CD=CB$ . Now $ABCC'$ is a quadrilateral whose one pair of opposite angles are equal and one pair of adjacent sides(not including one of the equal angles) are equal. It is now easy to say that $AB=AC'=AD+DC'=AD+DC$

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#8 h Apr 7, 2021, 1:47 AM

Y by

We know that $AC^2=AB^2+BC^2-2AB\cdot BC\cos(60)$

We also know that $\angle D=120$ so $AC^2=AD^2+CD^2-2AD\cdot CD\cos(120)$ .

We equate these to get
$\begin{align*} AB^2+BC^2-2AB\cdot BC\cos(60) &= AD^2+CD^2-2AD\cdot CD\cos(120) \\ AB^2+BC^2-AB\cdot BC &= AD^2+CD^2+AD\cdot CD \\ AB^2+CD^2-AB\cdot CD &= AD^2+CD^2+AD\cdot CD \\ AB^2-AB\cdot CD &= AD^2+AD\cdot CD \\ AB^2-AD^2 &= (AB+AD)CD \\ (AB-AD)(AB+AD) &= (AB+AD)CD \\ AB-AD &= CD \end{align*}$ so $AB=AD+DC$ .

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#9 h Apr 7, 2021, 1:55 AM

Y by

Hmmmmmm. I think mine is the cleanest so far.

Select point $E$ on $\overline{AB}$ such that $\triangle BCE$ is equilateral. It suffices to prove that $AE=AD\implies\angle ADE=\angle AED$ . Let $\angle DCE=\theta$ . Because $\triangle DCE$ is isosceles, see have $\angle CDE=90^\circ-\tfrac{\theta}{2}$ which implies $\angle ADE=30^\circ+\tfrac{\theta}{2}$ . Now, since $\angle BCD=60^\circ+\theta$ , we have $\angle DAE=120^\circ-\theta\implies\angle AED=30^\circ+\tfrac{\theta}{2}$ . Done.

@below, rip. Also, the motivation for this was a problem in Awesome Math's 106 Geo Problems(i think); I will find the exact one later.

This post has been edited 3 times. Last edited by Pleaseletmewin, Apr 7, 2021, 1:59 AM

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#10 h Apr 7, 2021, 1:57 AM

Y by

Pleaseletmewin wrote:

Hmmmmmm. I think mine is the cleanest so far.

Select point $E$ on $\overline{AB}$ such that $\triangle BCE$ is equilateral. It suffices to prove that $AE=AD\implies\angle ADE=\angle AED$ . Let $\angle DCE=\theta$ . Because $\triangle DCE$ is isosceles, see have $\angle CDE=90^\circ-\tfrac{\theta}{2}$ which implies $\angle ADE=30^\circ+\tfrac{\theta}{2}$ . Now, since $\angle BCD=60^\circ+\theta$ , we have $\angle DAE=120^\circ-\theta\implies\angle AED=30^\circ+\tfrac{\theta}{2}$ . Done.

mine was the ugliest

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