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Quad formed by orthocenters has same area (all 7's!)
v_Enhance   37
N 36 minutes ago by lpieleanu
Source: USA January TST for the 55th IMO 2014
Let $ABCD$ be a cyclic quadrilateral, and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. Let $W$, $X$, $Y$ and $Z$ be the orthocenters of triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively. Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.
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v_Enhance
Apr 28, 2014
lpieleanu
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Find the length of diagonal AC
Amir Hossein   14
N Mar 29, 2025 by ClassyPeach
Source: Bulgaria JBMO TST 2018, Day 1, Problem 1
In the quadrilateral $ABCD$, we have $\measuredangle BAD = 100^{\circ}$, $\measuredangle BCD = 130^{\circ}$, and $AB=AD=1$ centimeter. Find the length of diagonal $AC$.
14 replies
Amir Hossein
Jun 25, 2018
ClassyPeach
Mar 29, 2025
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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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Amir Hossein
5452 posts
Amir Hossein
#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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sunken rock
4384 posts
sunken rock
#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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Amir Hossein
5452 posts
Amir Hossein
#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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MeineMeinung
68 posts
MeineMeinung
#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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adityaguharoy
4655 posts
adityaguharoy
#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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Memmedov_Elvin
24 posts
Memmedov_Elvin
#7 h  Jun 25, 2018, 12:51 PM • 1 Y
Y by Adventure10
Answer is 1 cm
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LOM
84 posts
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
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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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PROF65
2016 posts
PROF65
#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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sunken rock
4384 posts
sunken rock
#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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biomathematics
2567 posts
biomathematics
#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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Steve12345
618 posts
Steve12345
#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
5047 posts
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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Illuzion
211 posts
Illuzion
#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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ClassyPeach
4 posts
ClassyPeach
#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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