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2004 IMO Problems/Problem 5

Revision as of 15:02, 8 February 2024 by Szhangmath (talk | contribs) (→‎Solution)

Problem

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.\]

Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP.$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Let $K$ be the intersection of $AC$ and $BE$, let $L$ be the intersection of $AC$ and $DF$,

$\angle PBC=\angle DBA$, so $AD=CE$, and $DE//AC$. $\angle PDC=\angle BDA$, so $AB=CF$, and $AC//BF$. 

$\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL$, so $\triangle PKL$ is an isosceles triangle.
Since $AC//BF$, so $\triangle PBF$ and $\triangle PDE$ are isosceles triangles. So $P$ is on the angle bisector oof $BF$, since $ABFC$ is 
an isosceles trapezoid, so $P$ is also on the perpendicular bisector of $AC$. So $PA=PC$.


[asy] import graph; size(13.98cm); 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 = -6.22, xmax = 7.76, ymin = -6.56, ymax = 6.3; /* image dimensions */ 

/* draw figures */

draw(circle((0.,0.), 4.), linewidth(2.)); draw((-3.9904248302051744,0.276603822247635)--(-1.2649110640673522,-3.794733192202055), linewidth(2.)); draw((-1.2649110640673522,-3.794733192202055)--(0.28102660741773866,-3.990115793548262), linewidth(2.)); draw((0.28102660741773866,-3.990115793548262)--(3.9893832569337877,0.2912408441416967), linewidth(2.)); draw((-3.9904248302051744,0.276603822247635)--(0.28102660741773866,-3.990115793548262), linewidth(2.)); draw((-3.9904248302051744,0.276603822247635)--(3.797959075020809,-1.2551919631941093), linewidth(2.)); draw((0.28102660741773866,-3.990115793548262)--(0.7148182881712134,3.935611110729308), linewidth(2.)); draw((3.9893832569337877,0.2912408441416967)--(0.7148182881712134,3.935611110729308), linewidth(2.) + linetype("4 4")); draw((-3.9904248302051744,0.276603822247635)--(3.9893832569337877,0.2912408441416967), linewidth(2.)); draw((-1.2649110640673522,-3.794733192202055)--(0.4665755573598317,-0.5999855308790458), linewidth(2.)); draw((3.9893832569337877,0.2912408441416967)--(0.4665755573598317,-0.5999855308790458), linewidth(2.)); draw((0.28102660741773866,-3.990115793548262)--(3.797959075020809,-1.2551919631941093), linewidth(2.)); draw((-1.2649110640673522,-3.794733192202055)--(3.9893832569337877,0.2912408441416967), linewidth(2.)); draw((-3.9904248302051744,0.276603822247635)--(0.7148182881712134,3.935611110729308), linewidth(2.)); draw((3.797959075020809,-1.2551919631941093)--(3.9893832569337877,0.2912408441416967), linewidth(2.) + linetype("2 2")); 

/* dots and labels */

dot((-1.2649110640673522,-3.794733192202055),dotstyle); label("$A$", (-1.64,-4.2), NE * labelscalefactor); dot((-3.9904248302051744,0.276603822247635),linewidth(4.pt) + dotstyle); label("$D_{2}$", (-4.52,0.1), NE * labelscalefactor); dot((0.7148182881712134,3.935611110729308),linewidth(4.pt) + dotstyle); label("$E$", (0.8,4.1), NE * labelscalefactor); dot((3.9893832569337877,0.2912408441416967),linewidth(4.pt) + dotstyle); label("$D$", (4.06,0.46), NE * labelscalefactor); dot((0.28102660741773866,-3.990115793548262),linewidth(4.pt) + dotstyle); label("$B$", (0.2,-4.46), NE * labelscalefactor); dot((3.797959075020809,-1.2551919631941093),linewidth(4.pt) + dotstyle); label("$F$", (4.04,-1.42), NE * labelscalefactor); dot((0.4665755573598317,-0.5999855308790458),linewidth(4.pt) + dotstyle); label("$P$", (0.54,-0.44), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 

/* end of picture */
[/asy]

~szhangmath

See Also

2004 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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