Been 24 hours so hopefully discussion is allowed.

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

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 $BC$?"

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!

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.

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 $CI=JB$

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.

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

Do you have figure this problem?

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.

wa3rrrmath wrote: Do you have figure this problem?

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 $CI = JB$ $\blacksquare$

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.

[asy][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][/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 $IC=JB$.

My solution:

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.

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 $\triangle EBJ\cong\triangle EIC.$ $\square$