Find the length of diagonal AC

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Find the length of diagonal AC

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Source: Bulgaria JBMO TST 2018, Day 1, Problem 1

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#1 h Jun 25, 2018, 8:24 AM • 3 Y

Y by Illuzion, Adventure10, Mango247

In the quadrilateral $ABCD$ , we have $\measuredangle BAD = 100^{\circ}$ , $\measuredangle BCD = 130^{\circ}$ , and $AB=AD=1$ centimeter. Find the length of diagonal $AC$ .

This post has been edited 1 time. Last edited by Amir Hossein, Jun 25, 2018, 9:07 AM
Reason: Fixed.

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#3 h Jun 25, 2018, 9:06 AM • 3 Y

Y by Adventure10, Mango247, AylyGayypow009

Amir Hossein wrote:

In the quadrilateral $ABCD$ , we have $\measuredangle BAD = 100^{\circ}$ , $\measuredangle BCD = 130^{\circ}$ , and $AD=1$ centimeter. Find the length of diagonal $AC$ .

Not enough data given!!

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#4 h Jun 25, 2018, 9:07 AM • 2 Y

Y by Adventure10, Mango247

sunken rock wrote:

Not enough data given!!

How about now? Thanks, fixed.

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#5 h Jun 25, 2018, 9:18 AM • 2 Y

Y by Adventure10, Mango247

Amir Hossein wrote:

In the quadrilateral $ABCD$ , we have $\measuredangle BAD = 100^{\circ}$ , $\measuredangle BCD = 130^{\circ}$ , and $AB=AD=1$ centimeter. Find the length of diagonal $AC$ .

I think it is still not enough. If we fix $A,B,D$ , the point $C$ can be any point on some circle.

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#6 h Jun 25, 2018, 11:47 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

MeineMeinung wrote:

Amir Hossein wrote:

In the quadrilateral $ABCD$ , we have $\measuredangle BAD = 100^{\circ}$ , $\measuredangle\color{red} {BCD} = 130^{\circ}$ , and $AB=AD=1$ centimeter. Find the length of diagonal $AC$ .

I think it is still not enough. If we fix $A,B,D$ , the point $C$ can be any point on some circle.

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#7 h Jun 25, 2018, 12:51 PM • 1 Y

Y by Adventure10

Answer is 1 cm

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LOM

#8 h Jun 25, 2018, 12:53 PM • 2 Y

Y by adityaguharoy, Adventure10

yes I think I is 1cm

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LOM

#9 h Jun 25, 2018, 12:59 PM • 4 Y

Y by Amir Hossein, adityaguharoy, Adventure10, Mango247

If you Stretch the altitude of ABC triangle to cut BC and call E new point ADCE is Cyclic quadrilateral and AEC=ACD , AED=ADC and we know AEC=AED so AD=AC=1

This post has been edited 1 time. Last edited by LOM, Jun 25, 2018, 1:01 PM
Reason: Fix

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#10 h Jun 25, 2018, 2:25 PM • 4 Y

Y by Amir Hossein, adityaguharoy, AlastorMoody, Adventure10

let $O$ the circumcircle of $BCD$ , we have $\angle BOC=100^\circ$ hence $BCAO$ are cyclic but $O,A$ li e on the $BC$ -bisector then $A=O$ implies $AC=1$

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#11 h Jun 27, 2018, 7:35 PM • 3 Y

Y by Amir Hossein, adityaguharoy, Adventure10

Draw the circle $C(A,AB)$ ; any point $X$ onto the bigger arc $\stackrel{\frown}{BD}$ has $m(\widehat{BXD})=50^\circ$ , that is, $XBCD$ is cyclic, wherefrom $AC=AB$ .

Best regards,
sunken rock

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#12 h Dec 4, 2018, 5:26 PM • 2 Y

Y by Adventure10, Mango247

We have

$\angle ABC + \angle CDA = 130^{\circ} = \angle BCD$ so
$(\angle ABC - \angle BCA ) + (\angle CDA - \angle ACD) = 0$ But the LHS has the same sign as that of
$(AC - AB) + (AC - AD) = 2(AC-1)$ so that $AC = 1$ .

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#13 h Dec 4, 2018, 5:43 PM • 3 Y

Y by sketchcomedyrules, Adventure10, Rotten_

We can also denote some $A_{1}$ such that $A_{1}BA$ is isosceles with angle $BA_{1}D$ =50° Then notice that $A_{1}BCD$ is cyclic and $A_{1}BD$ =90° and then get that $A$ is the center of the circumscribed circle. From here it is obvious that $AC$ is the radius $AC=AD=1$

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Pirkuliyev Rovsen

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Pirkuliyev Rovsen

#14 h Dec 4, 2018, 6:45 PM • 4 Y

Y by adityaguharoy, Adventure10, Mango247, Rotten_

This is from the competition of the Tournament Of Towns -1984-85

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#15 h Dec 25, 2018, 5:35 PM • 2 Y

Y by Adventure10, Mango247

WLOG let $BC>CD$ (the case $BC<CD$ is analogous). Let the perpendicular bisector of $BD$ intersect $BC$ at $L$ (this is due to $BC>CD$ in $\triangle BCD$ ). Since $\triangle BAD$ is isosceles, the perpendicular bisector will pass through $A$ . We now have $DL=BL$ and $AD=AB$ , so $\triangle ABL \cong \triangle ADL$ . Also, $\angle BAL=50$ and if we let $\angle ABC=\phi$ , we will get $\angle BLA=130-\phi$ , but $\angle ADC=130-\phi$ too, so $ALCD$ is cyclic. Now we have $\angle ABC=\angle ADL=\angle ACL$ , so $\triangle ABC$ is isosceles and $AC=AB=1$ . Now in the case $BC=CD$ we directly get that $ABCD$ is a deltoid, so $\angle ADC=65$ , while $\angle ACD=130/2=65$ and so $AC=AD=1$ .

This post has been edited 2 times. Last edited by Illuzion, Dec 25, 2018, 5:39 PM
Reason: missing case

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#16 h Mar 29, 2025, 12:29 PM

Y by

Because of limited symbols let's denote (/_) angle
/_BAD + 2× /_ BCD =360°. So A is a center and AB is a radius of a circle that C also lies on that circle. So AC=1

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