309. Beautiful problem ! IRAN Selection Test for IMO, 2004.
by Virgil Nicula, Aug 14, 2011, 12:06 AM
Very nice Darij Grinberg's post !
Lemma - Gergonne theorem on pedal triangles. Let
in the plane of 
the absolute value 
of the power of w.r.t. the circumcircle 
of
the orthogonal projections 
, 
, 
of on 
, 
, 
the area 
of 
. The area 
of 
can be computed using formula 
.
Proof. Let
a.s.o. I"ll use the well-known identity 
. We have 
and 
. Hence and lie on the
circle with diameter![$[BM]$](//latex.artofproblemsolving.com/b/8/d/b8df50365ac4274a59b15564d2715bbc99693b79.png)
. Thus, 
. Similarly, and lie on the circle with diameter ![$[AM]$](//latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
. Thus, 
. But 
and since 
, 
, 
and
are concyclic, we have 
. Thus, 
. Hence 
. Now by
another famous area formula for triangles, the area of is 
. Hence, 
. Since and lie on the circle with
diameter , the circumcircle of 
has diameter and thus, after the extended sine law, we get 
, or with other words, 
.
. Therefore, 
. On the other hand, 
. Therefore,
. By the sine law in triangle BMA', we have 
. Thus,
. Now, 
, but the concyclicity of , , and yields 
,
so that
. So we get 
, so that 
.
PP. Suppose that is a point inside of a triangle 
with the inradius 
. Denote 
. Prove that 
.
Proof. We know that
is the absolute value of the power of w.r.t. the circumcircle of triangle . Hence, the inequality we have to prove can be rewritten as
or 
. Now, let , , be the orthogonal projections of on the sides , , 
and let be the area of triangle
and the area of triangle 
. The Gergonne's theorem (see upper lemma) on pedal triangles states that . But on the other hand, if 
is circumradius of , then
. Thus, 
and 
. and lie on the circle with
diameter
. Hence, 
and similarly, 
, 
. Hence, if we multiply the inequality in question
with 
, we get 
or 
.
By the formula for
we derived above, this becomes equivalent to 
. With other words, the radius of the circumcircle of is greater or equal to the inradius of
. But this is known - The radius of any circle which cuts or touches all three sidelines of a triangle is greater or equal to the inradius of the triangle, with equality iff the
circle coincides with the incircle of the triangle. Hence, our inequality is proven. Equality holds only if the circumcircle of coincides with the incircle of , i. e. if is
the incenter of . Notably, the inequality we have proven is a generalization of the well-known 
inequality (just let be the circumcenter of ).
Remark. Another proof of
can obtain by the 
-point inequality from here post #1 to 
and 
in the circumcircle of .
Lemma - Gergonne theorem on pedal triangles. Let
in the plane of 
the absolute value 
of the power of w.r.t. the circumcircle 
of
the orthogonal projections 
, 
, 
of on 
, 
, 
the area 
of 
. The area 
of 
can be computed using formula 
.Proof. Let
a.s.o. I"ll use the well-known identity 
. We have 
and 
. Hence and lie on thecircle with diameter
![$[BM]$](http://latex.artofproblemsolving.com/b/8/d/b8df50365ac4274a59b15564d2715bbc99693b79.png)
. Thus, 
. Similarly, and lie on the circle with diameter ![$[AM]$](http://latex.artofproblemsolving.com/1/f/9/1f9b22599237fb6240a50b5f75e8f6ced1292374.png)
. Thus, 
. But 
and since 
, 
, 
and
are concyclic, we have 
. Thus, 
. Hence 
. Now byanother famous area formula for triangles, the area of
is 
. Hence, 
. Since and lie on the circle withdiameter
, the circumcircle of 
has diameter and thus, after the extended sine law, we get 
, or with other words, 
.
. Therefore, 
. On the other hand, 
. Therefore,
. By the sine law in triangle BMA', we have 
. Thus,
. Now, 
, but the concyclicity of , , and yields 
,so that
. So we get 
, so that 
.PP. Suppose that
is a point inside of a triangle 
with the inradius 
. Denote 
. Prove that 
.Proof. We know that
is the absolute value of the power of w.r.t. the circumcircle of triangle . Hence, the inequality we have to prove can be rewritten as
or 
. Now, let , , be the orthogonal projections of on the sides , , 
and let be the area of triangle and
the area of triangle 
. The Gergonne's theorem (see upper lemma) on pedal triangles states that . But on the other hand, if 
is circumradius of , then
. Thus, 
and 
. and lie on the circle withdiameter
. Hence, 
and similarly, 
, 
. Hence, if we multiply the inequality in question with 
, we get 
or 
.By the formula for
we derived above, this becomes equivalent to 
. With other words, the radius of the circumcircle of is greater or equal to the inradius of . But this is known - The radius of any circle which cuts or touches all three sidelines of a triangle is greater or equal to the inradius of the triangle, with equality iff thecircle coincides with the incircle of the triangle. Hence, our inequality is proven. Equality holds only if the circumcircle of
coincides with the incircle of , i. e. if isthe incenter of
. Notably, the inequality we have proven is a generalization of the well-known 
inequality (just let be the circumcenter of ).Remark. Another proof of
can obtain by the 
-point inequality from here post #1 to 
and 
in the circumcircle of . This post has been edited 34 times. Last edited by Virgil Nicula, Nov 20, 2015, 11:10 AM
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