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MortemEtInteritum

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MortemEtInteritum

#1 h Mar 12, 2021, 10:18 PM

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Let $ABCD$ be a trapezoid with $AB$ parallel to $CD$ , $|AB|>|CD|$ , and equal edges $|AD|=|BC|$ . Let $I$ be the center of the circle tangent to lines $AB$ , $AC$ and $BD$ , where $A$ and $I$ are on opposite sides of $BD$ . Let $J$ be the center of the circle tangent to lines $CD$ , $AC$ and $BD$ , where $D$ and $J$ are on opposite sides of $AC$ . Prove that $|IC|=|JB|$ .

This post has been edited 3 times. Last edited by MortemEtInteritum, Mar 13, 2021, 3:58 AM

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MortemEtInteritum

1332 posts

MortemEtInteritum

#2 h Mar 12, 2021, 10:21 PM

Y by

Been 24 hours so hopefully discussion is allowed.
Solution

@below Yes, my mistake. Edit: Nope, it wasn't that, but it was indeed wrong originally. Should be right now (hopefully).

This post has been edited 2 times. Last edited by MortemEtInteritum, Mar 12, 2021, 11:53 PM

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779 posts

#3 h Mar 12, 2021, 11:31 PM

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I didn't take the olympiad, but shouldn't the problem say "Let $I$ be the center of the circle tangent to lines $\mathbf{D}B,AC,$ and $BC$ ?"

This post has been edited 1 time. Last edited by amuthup, Mar 12, 2021, 11:31 PM

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#4 h Mar 15, 2021, 10:25 PM

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MortemEtInteritum wrote:

Let $ABCD$ be a trapezoid with $AB$ parallel to $CD$ , $|AB|>|CD|$ , and equal edges $|AD|=|BC|$ . Let $I$ be the center of the circle tangent to lines $AB$ , $AC$ and $BD$ , where $A$ and $I$ are on opposite sides of $BD$ . Let $J$ be the center of the circle tangent to lines $CD$ , $AC$ and $BD$ , where $D$ and $J$ are on opposite sides of $AC$ . Prove that $|IC|=|JB|$ .

I am not sure if the problem is formulated correctly, could you please double-check (the part about A & I being on different sides of BD)? Thanks!

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MortemEtInteritum

1332 posts

MortemEtInteritum

#5 h Mar 15, 2021, 11:14 PM

Y by

bora_olmez wrote:

MortemEtInteritum wrote:

Let $ABCD$ be a trapezoid with $AB$ parallel to $CD$ , $|AB|>|CD|$ , and equal edges $|AD|=|BC|$ . Let $I$ be the center of the circle tangent to lines $AB$ , $AC$ and $BD$ , where $A$ and $I$ are on opposite sides of $BD$ . Let $J$ be the center of the circle tangent to lines $CD$ , $AC$ and $BD$ , where $D$ and $J$ are on opposite sides of $AC$ . Prove that $|IC|=|JB|$ .

I am not sure if the problem is formulated correctly, could you please double-check (the part about A & I being on different sides of BD)? Thanks!

I do believe it is correct now, although it was originally posted with a few errors. Keep in mind the circle is tangent to those lines, not the line segments.

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1069 posts

gghx

#6 h Mar 16, 2021, 1:14 AM • 1 Y

Y by alibehzadi

Essentially, the problem should be phrased this way:
Let $ABCD$ be a isosceles trapezium with $CD//AB$ , $CD<AB$ , $AD=BC$ . Let $AC\cap BD=O$ . Let $I$ be the excenter of $AOB$ opposite $A$ And $J$ be the excenter of $DCO$ opposite $D$ . Show that $JB=IC$ .

In that case, here is an easy solution.
Note that $OI$ , $OJ$ are both angle bisectors of $COB$ And thus $O,I,J$ are collinear.
Now, $\angle IOB=\frac{1}{2}(180^\circ-\angle AOB=\angle OBA$ .
Thus, $\angle IOB=x \implies \angle OBA=x \implies \angle OBI=90-\frac{1}{2}x=\angle OIB$ . Hence, $OB=OI$ . Similarly, $OJ=OC$ . Combined with the fact that $\angle COI=\angle JOB$ , we have $\triangle COI \equiv \triangle JOB$ and thus $CI=JB$

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#7 h Apr 3, 2021, 7:45 AM • 1 Y

Y by I_am_human

Solution with p_square and Arwen713. Let $E=AC\cap BD$ . Now, we have that $\angle CEB=2\angle EDC\implies \angle JEC=2\angle JDC$ and also $\angle CJD=90-\angle EDC=\frac{1}{2}\angle CED$ . Thus, $E$ is the circumcenter of $(CJD)$ . Thus, $EJ=EC$ . Similarly, $EI=EB$ and $\angle JEB=\angle IEC$ . Thus, $\Delta IEC\cong \Delta BEJ$ . Thus, $BJ=IC$ and we are done.

This post has been edited 2 times. Last edited by Rg230403, Apr 3, 2021, 6:05 PM

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#8 h Apr 6, 2021, 7:59 PM • 1 Y

Y by george_54

Let X be the reflection of B across IJ. Now, just note that JIXC is a isosceles trapezoid and BJ=XJ=IC as needed

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#9 h Apr 11, 2021, 3:18 PM

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Do you have figure this problem?

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MortemEtInteritum

1332 posts

MortemEtInteritum

#10 h Apr 12, 2021, 6:03 AM • 3 Y

Y by Mango247, Mango247, Mango247

wa3rrrmath wrote:

Do you have figure this problem?

I have 0 Asy skill, so I didn't include a figure. The original problem didn't have one either.

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1585 posts

#11 h Apr 12, 2021, 10:04 AM • 1 Y

Y by Mango247

wa3rrrmath wrote:

Do you have figure this problem?

Attachments:

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#12 h Apr 13, 2021, 7:49 AM • 2 Y

Y by Mango247, Mango247

Let $O$ be the point of intersection of lines $OB$ and $AC$
Points $J$ and I both lie in the angle bisector of $\angle{COB}$ (Since both circles are tangent to lines $CO$ and $OB$ ). Now, $180-\angle{COB}=\angle{AOB}$ , Since $\triangle{AOB}$ is an isosceles triangle.
$\begin{align*} 180-\angle{COB} &= \angle{AOB} \\ 180-2\angle{COJ} &= \angle{AOB} \\ 180-2\angle{OBA} &= \angle{AOB} \\ 180-2\angle{COJ} &= 180-2\angle{OBA} \\ \angle{COJ} &= \angle{OBA} \\ \end{align*}$ Hence it follows that $OJ$ is parallel to $AB$ and to $DC$ .
Note that $CJ$ is the angle bisector of the exterior angle of $\angle{DCO}$ . Since $OJ$ is parallel to $DC$ , we have that $\angle{OJC}=\angle{OCJ}$ and hence $\triangle{OCJ}$ is an isosceles triangle, the same procedure works for showing that $\triangle{OIB}$ is also isosceles.
Hence $\triangle{COI}\cong \triangle{OJB}$ since $\angle{COI} = \angle{JOB}$ , $OB = OI$ and $OC = OJ$ . Consequently it follows that $CI = JB$ $\blacksquare$

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1574 posts

vvluo

#13 h May 18, 2021, 5:30 PM

Y by

~~vvluo~~ Coordinates have entered the chat.

Let $P$ be the intersection of $AC$ and $BD$ .
$I$ and $J$ are the $A$ -excenter of $APB$ and the $D$ -excenter of $DPC$ , respectively.
The $P$ -external bisector in both triangles is the line through $P$ parallel to $AB$ and $CD$ , so the intersection of that line withis the $A$ -internal bisector and the $D$ -internal bisector in their respective triangles are $I$ and $J$ , respectively.

~~We then discover our diagram is small and poor and quit using anything resembling synthetics.~~

By the properties of isosceles trapezoid $ABCD$ , we set the $x$ -axis to the line through $P$ parallel to $AB$ and $CD$ , and work out $A=(-a,ak)$ , $B=(a,ak)$ , $C=(d, -dk)$ , and $D=(-d,-dk)$ for some positive reals $a,d,k$ while $P$ is the origin.
We can bash out the slope of the bisectors from $A$ and $D$ by reverse solving the double angle tangent formula.
This slope is $-\frac{\sqrt{k^2 +1} -1}{k}$ and $\frac{\sqrt{k^2 +1} -1}{k}$ for $A$ and $D$ respectively.
The rest of the problem is just intersecting such lines with the $x$ -axis and using the distance formula, finishing with elementary algebraic manipulations.

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#14 h Jun 12, 2021, 10:43 AM • 1 Y

Y by Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.635255672046435, xmax = 11.863664117724507, ymin = -4.20704291769339, ymax = 10.440265830190652; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((-5,0)--(-3.6230465173026754,5.342454659878084), linewidth(0.4) + blue); draw((-3.6230465173026754,5.342454659878084)--(1.0436198161201171,5.342454659878084), linewidth(0.4) + blue); draw((1.0436198161201171,5.342454659878084)--(2.4205732988174433,0), linewidth(0.4) + blue); draw((2.4205732988174433,0)--(-5,0), linewidth(0.4) + blue); draw((-5,0)--(1.0436198161201171,5.342454659878084), linewidth(0.4) + blue); draw((-3.6230465173026754,5.342454659878084)--(2.4205732988174433,0), linewidth(0.4) + blue); draw(circle((3.6624077946841904,3.2798287785650975), 3.2798287785650975), linewidth(0.4) + linetype("4 4") + red); draw((1.0436198161201171,5.342454659878084)--(1.4901514016905097,5.73718080458862), linewidth(0.4) + blue); draw((2.4205732988174433,0)--(3.66240779468419,0), linewidth(0.4) + blue); draw((-1.2897133505912788,3.279828778565098)--(3.6624077946841904,3.2798287785650975), linewidth(0.4) + blue); draw((1.0436198161201171,5.342454659878084)--(3.6624077946841904,3.2798287785650975), linewidth(0.4) + blue); draw((2.4205732988174433,0)--(1.8245874044960655,3.279828778565098), linewidth(0.4) + blue); draw(circle((1.8245874044960655,3.279828778565098), 2.0626258813129863), linewidth(0.4) + linetype("4 4") + red); draw((1.0436198161201171,5.342454659878084)--(1.8245874487410427,5.342454659878084), linewidth(0.4) + blue); draw(circle((-1.2897133505912788,3.279828778565098), 4.952121145275469), linewidth(0.4) + linetype("4 4") + qqwuqq); draw((3.6624077946841904,3.2798287785650975)--(2.4205732988174433,0), linewidth(0.4) + blue); draw((3.6624077946841904,3.2798287785650975)--(-5,0), linewidth(0.4) + blue); draw(circle((-1.2897133505912788,3.279828778565098), 3.1143007550873447), linewidth(0.4) + linetype("4 4") + qqwuqq); draw((-3.6230465173026754,5.342454659878084)--(1.8245874044960655,3.279828778565098), linewidth(0.4) + blue); /* dots and labels */ dot((-5,0),dotstyle); label("$A$", (-4.891171068704102,0.2714507642136014), NE * labelscalefactor); dot((2.4205732988174433,0),dotstyle); label("$B$", (2.5378596271651874,0.2714507642136014), NE * labelscalefactor); dot((1.0436198161201171,5.342454659878084),dotstyle); label("$C$", (1.1416233616294698,5.592955021538379), NE * labelscalefactor); dot((-3.6230465173026754,5.342454659878084),linewidth(4pt) + dotstyle); label("$D$", (-3.5212788836501905,5.540266860574768), NE * labelscalefactor); dot((-1.2897133505912788,3.279828778565098),linewidth(4pt) + dotstyle); label("$E$", (-1.1766557207694572,3.485428582993913), NE * labelscalefactor); dot((3.6624077946841904,3.2798287785650975),linewidth(4pt) + dotstyle); label("$I$", (3.7760314098100687,3.485428582993913), NE * labelscalefactor); dot((1.8245874044960655,3.279828778565098),linewidth(4pt) + dotstyle); label("$J$", (1.9319457760836496,3.485428582993913), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Nice and easy geo, pretty astonished that no one has posted something similar.

Let $E$ be the intersection of $AC$ and $BD$ .

Obviously we must have that $ABCD$ must be a isosceles trapezoid.
By the given definitions of $J$ and $I$ , we must have that $J$ is the $D$ -excenter of $DEC$ and that $I$ is the $A$ -excenter $AEB$ .

Since we have that $ABCD$ is isosceles we must have that $\angle BAC = \angle ACD = \angle CDB = \angle DBA = \alpha$ .

Now notice that the points $E,J$ and $I$ must be collinear, since we have that $\alpha = \angle JEB = \angle IEB$ .

Now notice that $\angle AIB = 90-\alpha = \frac{1}{2}\angle AEB$ , and that $\angle IAB = \frac{1}{2}\angle IEB$ , which implies that $E$ is the circumcenter of $ABI$ .
Similarly we see that $E$ is the circumcenter of $DCJ$ .

Thus we have that $EJ=EC$ and that $EB=EI$ and that $\angle JEB = \angle IEC$ , thus by SAS we have that $\triangle EJB \cong \triangle ECI$ .
This implies that $IC=JB$ .

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#15 h Aug 16, 2021, 9:27 AM • 1 Y

Y by xiaoxun

My solution:
proof

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#16 h Dec 19, 2021, 7:07 AM

Y by

Let $AC$ and $BD$ meet at $E$ . we will prove $ECI$ and $EJB$ are congruent.
$\angle CEI = \angle CAB = 2\angle IAE$ ---> $IAE$ is isosceles ---> $EI = EA = EB$
$\angle BEJ = \angle BDC = 2\angle EDJ$ ---> $EDJ$ is isosceles ---> $EJ = ED = EC$
so now we have $IE = BE$ , $\angle IEC = \angle BEJ$ and $EC = EJ$ and with these we can prove $ECI$ and $EJB$ are congruent.
we're Done.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Mar 4, 2022, 8:40 AM

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#17 h Jan 28, 2022, 10:49 PM • 1 Y

Y by centslordm

Let $E=\overline{AC}\cap\overline{BD}$ and let $X$ be on $\overline{AB}$ beyond $B$ and note that $E$ lies on $\overline{IJ}$ so $\angle JEB=\angle CEI.$ Since $\overline{IJ}\parallel\overline{AB},$ $\angle BIE=\angle IBX=\angle EBI$ and $EI=EB.$ Similarly, $EJ=EC$ so $\triangle EBJ\cong\triangle EIC.$ $\square$

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lifeismathematics

1188 posts

lifeismathematics

#18 h Nov 28, 2022, 10:48 AM

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cute

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DensSv

#19 h Nov 11, 2024, 2:41 PM

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Sol.

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