folding a rectangle, ratio of segments wanted

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Source: V.A. Yasinsky Geometry Olympiad 2017 VIII-IX p1 [Ukrane]

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#1 h Sep 23, 2018, 8:59 AM • 2 Y

Y by Adventure10, Mango247

Rectangular sheet of paper $ABCD$ is folded as shown in the figure. Find the rato $DK: AB$ , given that $C_1$ is the midpoint of $AD$ .
//cdn.artofproblemsolving.com/images/0/1/d/01d3260c80145a1b00c80050ba1587430027e900.png

//cdn.artofproblemsolving.com/images/0/1/d/01d3260c80145a1b00c80050ba1587430027e900.png

This post has been edited 1 time. Last edited by parmenides51, Oct 11, 2020, 10:53 PM

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#2 h Sep 23, 2018, 9:22 AM • 2 Y

Y by Adventure10, Mango247

$\frac{1}{2}=\sin(ABC_1)=\sin(KC_1D)=\frac{DK}{KC_1}=\frac{DK}{AB-DK}$
$\to \frac{DK}{AB}=1/3$

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Pinko

#3 h Sep 23, 2018, 11:43 AM • 1 Y

Y by Adventure10

As $BC=BC_1$ we have that $\frac{AC_1}{BC_1}=\frac{0.5AD}{BC}=\frac{0.5BC}{BC}=\frac{1}{2}$ . Hence $\angle ABC_1=30^\circ$ .

Having that $\angle ABC_1=\angle DC_1K=30^\circ$ we get $sin 30^\circ =\frac{KD}{KC_1}=\frac{KD}{KC}=\frac{KD}{DC-KD}=\frac{KD}{AB-KD}$

Thus $AB-DK=2DK$ , $AB=3DK$ and $\frac{DK}{AB}=\frac{DK}{3DK}=\frac{1}{3}$ .

This post has been edited 1 time. Last edited by Pinko, Sep 23, 2018, 11:43 AM

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#4 h Nov 9, 2018, 1:17 PM • 1 Y

Y by Adventure10

It's clear that, $\Delta BC_1C$ is equilateral,
If $\angle KC_1D =x \implies \angle C_1CD=45^{\circ}-\frac {x}{2}=30^{\circ} \implies \boxed {x=30^{\circ}}$
Hence, $\frac {DK}{KC_1}=\frac {DK}{CK}=\frac {1}{2} \implies \frac {CK}{DK}=2 \implies DK:AB=1:3$

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#5 h Nov 9, 2018, 3:22 PM • 2 Y

Y by Adventure10, Mango247

A similar kind of problem appeared in Indian RMO 1990 P3, here,
https://artofproblemsolving.com/community/c6h56050p347212

This post has been edited 1 time. Last edited by AlastorMoody, Nov 9, 2018, 3:23 PM

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