273. The area of the triangle IOH. When I belongs to OH ?
by Virgil Nicula, May 6, 2011, 2:17 AM
Proposed problem 1. Let 
, 
, 
be the incenter, the circumcenter and the orthocenter respectively of a triangle 
with the
lengths of sides
, 
, 
and the inradius 
. Prove that the area of the triangle 
is ![$[IOH]=\frac {|(a-b)(b-c)(c-a)|}{8r}$](//latex.artofproblemsolving.com/3/4/0/340475e1bd4c2e44a32bcca7d717ef3578757ab8.png)
.
Proof. Let us use areal coordinates (normalized barycentric coordinates) w.r.t.
. Coordinates of the centroid, orthocenter and incenter are given by 
, 
, 
. Thus, 
![$\frac {[IGH]}{[ABC]}$](//latex.artofproblemsolving.com/1/7/7/177921dd409c662cb2ae9c8a79c69ef739783c62.png)
![$ 6p \cdot \frac {[IGH]}{[ABC]} = \sum_{\text{cyclic}} a(\cot A\cot C-\cot A\cot B)$](//latex.artofproblemsolving.com/2/6/b/26bede22c2f2a164f7aa5aefaeea26bff82438da.png)
. Using the well-known relations ![$ \cot A = \frac {b^2 + c^2 - a^2}{4[ABC]}$](//latex.artofproblemsolving.com/a/6/6/a66d356139b30f0f822afa50efedaad4e2a3bdfa.png)
, ![$\cot B = \frac {a^2 + c^2 - b^2}{4[ABC]}$](//latex.artofproblemsolving.com/2/1/9/2197daa0aefd2bf00f778a0f48beb1196fa72101.png)
,
![$\cot C = \frac {a^2 + b^2 - c^2}{4[ABC]}$](//latex.artofproblemsolving.com/d/9/c/d9c8dc410412baa096a79572dcc9ab140a9540b9.png)
obtain that ![$ \frac {48\cdot [IHG] \cdot [ABC]^2}{r} = \sum_{\text{cyclic}} a^4(c - b) + a^3(c^2 - b^2)$](//latex.artofproblemsolving.com/0/4/a/04ab510f52477f61a431b164101c5d2865e91afe.png)
![$48\cdot [IHG] \cdot [ABC]^2 = $](//latex.artofproblemsolving.com/d/8/0/d804e6228c38f02a2784b4c758968ccc09ca6336.png)
![$2rp[a^3(c - b) + b^3(a - c) + c^3(b - a)]$](//latex.artofproblemsolving.com/5/5/a/55a859bdd0706b4816a43d34d6a47a100daf4891.png)
![$\ [IHG]$](//latex.artofproblemsolving.com/9/f/8/9f82ab70801d1e882985569e27a46ec37284abbf.png)
![$ = \frac {a^3(c - b) + b^3(a - c) + c^3(b - a)}{24\cdot [ABC]}=$](//latex.artofproblemsolving.com/5/d/7/5d705345de886d0f7ede0021ef31603a43e5c68c.png)
.
Thus,![$2\cdot [IOH] = 3\cdot [IHG] \Longrightarrow$](//latex.artofproblemsolving.com/2/9/3/2937f13e635c642020fac5228562517c6e716db5.png)
![$[IOH] = \frac {|(a - b)(b - c)(c - a)|}{8r}$](//latex.artofproblemsolving.com/9/a/3/9a31976f0b4a91c6b7ce1b058c1d2e33d7bb6a06.png)
.
Application. If, 
, , , are the inradius, the circumradius, the circumcenter, the incenter and the
orthocenter respectively of the triangle, then ![$[OIH] \ge 4rR \sin\frac{|A-B|}{2} \sin \frac{|B-C|}{2} \sin \frac{|C-A|}{2}$](//latex.artofproblemsolving.com/0/d/d/0dd4d088c2b1c93449e68251f9b285539b99dffe.png)
.
Proof. From Mollweide formulas we have
a.s.o. 
. But 
. From the above proposed problem obtain that
![$[OIH]=\frac{|(a-b)(b-c)(c-a)|}{8r} \ \ (2)$](//latex.artofproblemsolving.com/c/e/7/ce7b985a8f8064fb063ad5d4ae363ed95b3ee1c1.png)
. Substituting 
and 
into the desired inequality gives 
, which is clearly true.
Proposed problem 2. Ascertain the nature of
such that the incenter 
, where is orthocenter and 
is centroid of .
Proof 1.
.
Proof 2. We will show that is isosceles. Well, assume it is not. According to the condition of the problem, the incenter , i. e. on the line 
,
where is the circumcenter of . The line 
bisects the 
. Hence 
. Similarly of course, 
and 
.
Now, as
we get 
. This readily yields that is equilateral and thus isosceles, contradicting with our assumption.
Proposed problem 3. Consider
with 
, 
, 
. Circle 
with center 
which lies outside 
is tangent to 
at the midpoint
of![$ [AB] $](//latex.artofproblemsolving.com/1/c/0/1c02f93ac05404f6709341e43643610476eed9a5.png)
and the circumcircle of 
. Circle 
with center 
which lies outside is tangent to 
at the midpoint of and the circumcircle of .
Circle
with center 
which lies outside is tangent to 
at the midpoint of and the circumcircle of . What is the area of 
?
Proof. Observe that is acute, 
and 
.
Therefore,
. Thus, ![$[YOZ]=\frac 12\cdot OY\cdot OZ\cdot\sin \widehat{YOZ}=$](//latex.artofproblemsolving.com/8/0/2/8020790538d14a2dc6a4f5600f612c0b5f1069a4.png)
![$\boxed{\ [YOZ]=\frac {(s-b)(s-c)\cdot S}{16r^2}\ }$](//latex.artofproblemsolving.com/0/b/5/0b57781cf10598277457b5dd0fc73d76d08091a8.png)
. Therefore, ![$[XYZ]=\sum [YOZ]=\frac {S}{16r^2}\cdot\sum (s-b)(s-c)=$](//latex.artofproblemsolving.com/8/8/3/8839725ff7c6fb5c1fb88f82ba47d4bb8bab11a6.png)
![$\boxed{\boxed {\ [XYZ]=\frac {s(4R+r)}{16}\ }}\ (*)$](//latex.artofproblemsolving.com/8/e/9/8e97bb1d505cc67017eacc83652d7756b7c30464.png)
. In conclusion, 
![$[XYZ]=\frac {21\cdot\left(4\cdot \frac {65}{8}+4\right)}{16}=$](//latex.artofproblemsolving.com/a/5/2/a525897a518ca8027961e04256e5f59cf9900073.png)
.
, 
, 
be the incenter, the circumcenter and the orthocenter respectively of a triangle 
with thelengths of sides
, 
, 
and the inradius 
. Prove that the area of the triangle 
is ![$[IOH]=\frac {|(a-b)(b-c)(c-a)|}{8r}$](http://latex.artofproblemsolving.com/3/4/0/340475e1bd4c2e44a32bcca7d717ef3578757ab8.png)
.Proof. Let us use areal coordinates (normalized barycentric coordinates) w.r.t.
. Coordinates of the centroid, orthocenter and incenter are given by 
, 
, 
. Thus, 
![$\frac {[IGH]}{[ABC]}$](http://latex.artofproblemsolving.com/1/7/7/177921dd409c662cb2ae9c8a79c69ef739783c62.png)
![$ 6p \cdot \frac {[IGH]}{[ABC]} = \sum_{\text{cyclic}} a(\cot A\cot C-\cot A\cot B)$](http://latex.artofproblemsolving.com/2/6/b/26bede22c2f2a164f7aa5aefaeea26bff82438da.png)
. Using the well-known relations ![$ \cot A = \frac {b^2 + c^2 - a^2}{4[ABC]}$](http://latex.artofproblemsolving.com/a/6/6/a66d356139b30f0f822afa50efedaad4e2a3bdfa.png)
, ![$\cot B = \frac {a^2 + c^2 - b^2}{4[ABC]}$](http://latex.artofproblemsolving.com/2/1/9/2197daa0aefd2bf00f778a0f48beb1196fa72101.png)
,![$\cot C = \frac {a^2 + b^2 - c^2}{4[ABC]}$](http://latex.artofproblemsolving.com/d/9/c/d9c8dc410412baa096a79572dcc9ab140a9540b9.png)
obtain that ![$ \frac {48\cdot [IHG] \cdot [ABC]^2}{r} = \sum_{\text{cyclic}} a^4(c - b) + a^3(c^2 - b^2)$](http://latex.artofproblemsolving.com/0/4/a/04ab510f52477f61a431b164101c5d2865e91afe.png)
![$48\cdot [IHG] \cdot [ABC]^2 = $](http://latex.artofproblemsolving.com/d/8/0/d804e6228c38f02a2784b4c758968ccc09ca6336.png)
![$2rp[a^3(c - b) + b^3(a - c) + c^3(b - a)]$](http://latex.artofproblemsolving.com/5/5/a/55a859bdd0706b4816a43d34d6a47a100daf4891.png)
![$\ [IHG]$](http://latex.artofproblemsolving.com/9/f/8/9f82ab70801d1e882985569e27a46ec37284abbf.png)
![$ = \frac {a^3(c - b) + b^3(a - c) + c^3(b - a)}{24\cdot [ABC]}=$](http://latex.artofproblemsolving.com/5/d/7/5d705345de886d0f7ede0021ef31603a43e5c68c.png)
.Thus,
![$2\cdot [IOH] = 3\cdot [IHG] \Longrightarrow$](http://latex.artofproblemsolving.com/2/9/3/2937f13e635c642020fac5228562517c6e716db5.png)
![$[IOH] = \frac {|(a - b)(b - c)(c - a)|}{8r}$](http://latex.artofproblemsolving.com/9/a/3/9a31976f0b4a91c6b7ce1b058c1d2e33d7bb6a06.png)
.Application. If
, 
, , , are the inradius, the circumradius, the circumcenter, the incenter and theorthocenter respectively of the triangle
, then ![$[OIH] \ge 4rR \sin\frac{|A-B|}{2} \sin \frac{|B-C|}{2} \sin \frac{|C-A|}{2}$](http://latex.artofproblemsolving.com/0/d/d/0dd4d088c2b1c93449e68251f9b285539b99dffe.png)
.Proof. From Mollweide formulas we have
a.s.o. 
. But 
. From the above proposed problem obtain that![$[OIH]=\frac{|(a-b)(b-c)(c-a)|}{8r} \ \ (2)$](http://latex.artofproblemsolving.com/c/e/7/ce7b985a8f8064fb063ad5d4ae363ed95b3ee1c1.png)
. Substituting 
and 
into the desired inequality gives 
, which is clearly true.Proposed problem 2. Ascertain the nature of
such that the incenter 
, where is orthocenter and 
is centroid of .Proof 1.
.Proof 2. We will show that
is isosceles. Well, assume it is not. According to the condition of the problem, the incenter , i. e. on the line 
,where
is the circumcenter of . The line 
bisects the 
. Hence 
. Similarly of course, 
and 
.Now, as
we get 
. This readily yields that is equilateral and thus isosceles, contradicting with our assumption.Proposed problem 3. Consider
with 
, 
, 
. Circle 
with center 
which lies outside 
is tangent to 
at the midpointof
![$ [AB] $](http://latex.artofproblemsolving.com/1/c/0/1c02f93ac05404f6709341e43643610476eed9a5.png)
and the circumcircle of 
. Circle 
with center 
which lies outside is tangent to 
at the midpoint of and the circumcircle of .Circle
with center 
which lies outside is tangent to 
at the midpoint of and the circumcircle of . What is the area of 
?Proof. Observe that
is acute, 
and 
.Therefore,
. Thus, ![$[YOZ]=\frac 12\cdot OY\cdot OZ\cdot\sin \widehat{YOZ}=$](http://latex.artofproblemsolving.com/8/0/2/8020790538d14a2dc6a4f5600f612c0b5f1069a4.png)
![$\boxed{\ [YOZ]=\frac {(s-b)(s-c)\cdot S}{16r^2}\ }$](http://latex.artofproblemsolving.com/0/b/5/0b57781cf10598277457b5dd0fc73d76d08091a8.png)
. Therefore, ![$[XYZ]=\sum [YOZ]=\frac {S}{16r^2}\cdot\sum (s-b)(s-c)=$](http://latex.artofproblemsolving.com/8/8/3/8839725ff7c6fb5c1fb88f82ba47d4bb8bab11a6.png)
![$\boxed{\boxed {\ [XYZ]=\frac {s(4R+r)}{16}\ }}\ (*)$](http://latex.artofproblemsolving.com/8/e/9/8e97bb1d505cc67017eacc83652d7756b7c30464.png)
. In conclusion, 
![$[XYZ]=\frac {21\cdot\left(4\cdot \frac {65}{8}+4\right)}{16}=$](http://latex.artofproblemsolving.com/a/5/2/a525897a518ca8027961e04256e5f59cf9900073.png)
.This post has been edited 25 times. Last edited by Virgil Nicula, Nov 22, 2015, 7:54 AM
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