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Difference between revisions of "2004 IMO Problems/Problem 5"

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{{solution}}
 
{{solution}}
  
−
Let <math>K</math>  be the intersection of <math>AC</math> and <math>BE</math>, let <math>L</math> be the intersection of <math>AC</math> and <math>DF</math>,
+
Assume <math>ABCD</math> is cyclic,
 +
let <math>K</math>  be the intersection of <math>AC</math> and <math>BE</math>, let <math>L</math> be the intersection of <math>AC</math> and <math>DF</math>,
 +
 
 +
<asy>
 +
size(6cm);
 +
draw(circle((0,0),7.07));
 +
draw((-3.7,-6)-- (3.7,-6));
 +
draw((-6.8,-2)-- (6.8,-2));
 +
draw((-5,5)-- (5,5));
 +
draw((-5,5)-- (-3.7,-6));
 +
draw((-5,5)-- (3.7,-6));
 +
draw((-5,5)-- (-6.8,-2));
 +
draw((-5,5)-- (6.8,-2));
 +
draw((5,5)-- (-3.7,-6));
 +
draw((5,5)-- (3.7,-6));
 +
draw((5,5)-- (-6.8,-2));
 +
draw((5,5)-- (6.8,-2));
 +
draw((-3.7,-6)-- (-6.8,-2));
 +
draw((-3.7,-6)-- (6.8,-2));
 +
draw((3.7,-6)-- (-6.8,-2));
 +
draw((3.7,-6)-- (6.8,-2));
 +
label("$A$", (-6.8,-2), SW);
 +
label("$B$", (-3.7,-6), SW);
 +
label("$F$", (3.7,-6), SE);
 +
label("$C$", (6.8,-2), E);
 +
label("$E$", (5,5), E);
 +
label("$D$", (-5,5), W);
 +
label("$P$", (0,-1.3), N);
 +
label("$K$", (-1.6,-1.5), E);
 +
label("$L$", (0.8,-1.5) );                       
 +
</asy>
  
 
<math>\angle PBC=\angle DBA</math>, so <math>AD=CE</math>, and <math>DE//AC</math>.
 
<math>\angle PBC=\angle DBA</math>, so <math>AD=CE</math>, and <math>DE//AC</math>.
 
<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
 
<math>\angle PDC=\angle BDA</math>, so <math>AB=CF</math>, and <math>AC//BF</math>.
 
  <math>\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
 
  <math>\angle PLK=\frac12(\overarc{AD}+\overarc{CF})=\frac12(\overarc{CE}+\overarc{AB})=\angle PKL</math>, so <math>\triangle PKL</math> is an isosceles triangle.
−
  Since <math>AC//BF</math>, so <math>\triangle PBF</math> and <math>\triangle PDE</math> are isosceles triangles. So <math>P</math> is on the angle bisector oof <math>BF</math>, since <math>ABFC</math> is  
+
  Since <math>AC//BF</math>, so <math>\triangle PBF</math> and <math>\triangle PDE</math> are isosceles triangles. So <math>P</math> is on the perpendicular bisector of <math>BF</math>, since <math>ABFC</math> is  
 
  an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
 
  an isosceles trapezoid, so <math>P</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
  
  
  
−
[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("<math>A</math>", (-1.64,-4.2), NE * labelscalefactor);
 
−
dot((-3.9904248302051744,0.276603822247635),linewidth(4.pt) + dotstyle);
 
−
label("<math>D_{2}</math>", (-4.52,0.1), NE * labelscalefactor);
 
−
dot((0.7148182881712134,3.935611110729308),linewidth(4.pt) + dotstyle);
 
−
label("<math>E</math>", (0.8,4.1), NE * labelscalefactor);
 
−
dot((3.9893832569337877,0.2912408441416967),linewidth(4.pt) + dotstyle);
 
−
label("<math>D</math>", (4.06,0.46), NE * labelscalefactor);
 
−
dot((0.28102660741773866,-3.990115793548262),linewidth(4.pt) + dotstyle);
 
−
label("<math>B</math>", (0.2,-4.46), NE * labelscalefactor);
 
−
dot((3.797959075020809,-1.2551919631941093),linewidth(4.pt) + dotstyle);
 
−
label("<math>F</math>", (4.04,-1.42), NE * labelscalefactor);
 
−
dot((0.4665755573598317,-0.5999855308790458),linewidth(4.pt) + dotstyle);
 
−
label("<math>P</math>", (0.54,-0.44), NE * labelscalefactor);
 
−
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 
−
/* end of picture */
 
−
[/asy]
 
 
~szhangmath
 
~szhangmath
  

Latest revision as of 22:28, 8 February 2024

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.

Assume $ABCD$ is cyclic, let $K$ be the intersection of $AC$ and $BE$, let $L$ be the intersection of $AC$ and $DF$,

[asy] size(6cm); draw(circle((0,0),7.07)); draw((-3.7,-6)-- (3.7,-6)); draw((-6.8,-2)-- (6.8,-2)); draw((-5,5)-- (5,5)); draw((-5,5)-- (-3.7,-6)); draw((-5,5)-- (3.7,-6)); draw((-5,5)-- (-6.8,-2)); draw((-5,5)-- (6.8,-2)); draw((5,5)-- (-3.7,-6)); draw((5,5)-- (3.7,-6)); draw((5,5)-- (-6.8,-2)); draw((5,5)-- (6.8,-2)); draw((-3.7,-6)-- (-6.8,-2)); draw((-3.7,-6)-- (6.8,-2)); draw((3.7,-6)-- (-6.8,-2)); draw((3.7,-6)-- (6.8,-2)); label("$A$", (-6.8,-2), SW); label("$B$", (-3.7,-6), SW); label("$F$", (3.7,-6), SE); label("$C$", (6.8,-2), E); label("$E$", (5,5), E); label("$D$", (-5,5), W); label("$P$", (0,-1.3), N); label("$K$", (-1.6,-1.5), E); label("$L$", (0.8,-1.5) ); [/asy]

$\angle PBC=\angle DBA$, 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 perpendicular bisector of $BF$, since $ABFC$ is 
an isosceles trapezoid, so $P$ is also on the perpendicular bisector of $AC$. So $PA=PC$.


~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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