Traditional Filipino cuisine
by cjquines0, Nov 15, 2016, 2:31 PM
Philippine MO 2010/2 wrote:
On a cyclic quadrilateral
, there is a point
on side
such that the triangle
and the quadrilateral
have equal perimeters and equal areas. Prove that two sides of
have equal lengths.






Solution
Let
have lengths
respectively. Because the perimeters of
and
are equal, we have
. We also have
.
Note that
. Then note that
and
.
Substituting both of these into the previous equation for areas, and applying the equation from the perimeters, gives us
, or
. Thus either
or
, which was what we wanted.





![$[CDP] = [ABCP] = [ABC] + [ACP]$](http://latex.artofproblemsolving.com/b/5/5/b559389cee552837439388bdb5cc0899f6d92815.png)
Note that
![$[ACP] = \frac{y}{x}[CDP]$](http://latex.artofproblemsolving.com/d/7/4/d741ec0442608f59097dbc28ae290753772c1a08.png)
![$[ABC] = \frac{1}{2}ab \sin \angle ABC$](http://latex.artofproblemsolving.com/0/c/8/0c8b6a8efcd43232ce1e0c35fe60e60e8dca867e.png)
![$[CDP] = \frac{1}{2}cx \sin \angle CDP = \frac{1}{2}cx \sin \angle ABC$](http://latex.artofproblemsolving.com/f/8/a/f8a1e8cf4e3b92ea4d2400956e5cc76883b6c3f5.png)
Substituting both of these into the previous equation for areas, and applying the equation from the perimeters, gives us




Tidbit
This is one of the more interesting problems that has surfaced in the PMO. Whether this is a compliment or not is for the reader to determine.
I am so hyped for the contest this Saturday! I’ve been training a lot with the other guy from our school. It’s been really exciting.
I am so hyped for the contest this Saturday! I’ve been training a lot with the other guy from our school. It’s been really exciting.