Difference between revisions of "2003 IMO Problems/Problem 4"
(Created page with "==Problem== Let <math>ABCD</math> be a cyclic quadrilateral. Let <math>P</math>, <math>Q</math>, and <math>R</math> be the feet of perpendiculars from <math>D</math> to lines...") |
|||
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
| − | {{ | + | Clearly <math>PQR</math> is the Simson Line and <math>APDQ</math>, <math>BPDR</math>, <math>CQDR</math> is cyclic. By angle chasing we have <math>\triangle DPQ\sim\triangle DBC</math>, <math>\triangle DQR\sim\triangle DAB</math>. Then by <math>PQ=QR</math> we have <math>\frac{DC}{CB}=\frac{DQ}{QP}=\frac{DQ}{QR}=\frac{DA}{AB}</math>. Rearranging and using the angle bisector theorem we are done. |
==See Also== | ==See Also== | ||
{{IMO box|year=2003|num-b=3|num-a=5}} | {{IMO box|year=2003|num-b=3|num-a=5}} | ||
Latest revision as of 03:07, 26 March 2024
Problem
Let 
be a cyclic quadrilateral. Let 
, 
, and 
be the feet of perpendiculars from 
to lines 
, 
, and 
, respectively. Show that 
if and only if the bisectors of angles 
and 
meet on segment 
.
Solution
Clearly 
is the Simson Line and 
, 
, 
is cyclic. By angle chasing we have 
, 
. Then by we have 
. Rearranging and using the angle bisector theorem we are done.
See Also
| 2003 IMO (Problems) • Resources | ||
| Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
| All IMO Problems and Solutions | ||
