281. Identity with R in an acute triangle (ILL - 1974).
by Virgil Nicula, May 26, 2011, 2:31 AM
PP. Let 
be an acute triangle with the circumcircle 
. Parallel through 
to 
cuts 
, 
at 
, 
respectively and define similarly the points
, 
and 
, 
. Then 
.
Proof. Sinus' theorem in
because 
.
Generalization. Let
be an interior point of 
with circumcircle . Consider the points 
, 
and
so that 
, 
, 
and 
. Then 
.
Proof. Let
be the orthogonal projections of onto , 
, . Denote the circumradius 
of 
. By Euler's theorem for and its pedal triangle
w.r.t. we get ![$\frac{[PYZ]}{[AA_1A_2]}=\frac{{\overline{PA_1}} \cdot \overline{PA_2}}{4{R_1}^2}$](//latex.artofproblemsolving.com/4/5/e/45e5b48f0a3a94c9cac8898d966283f625d5e75e.png)
. But 
gives ![$\frac{[ABC]}{[A_1B_1C_1]}=\frac{R^2}{{R_1}^2}$](//latex.artofproblemsolving.com/7/6/4/764e93bf51959ccbf8fcad571aa06ab4ea3eee8e.png)
![$\Longrightarrow \ \frac{[PYZ]}{[ABC]}=\frac{{\overline{PA_1}} \cdot \overline{PA_2}}{4R^2}$](//latex.artofproblemsolving.com/e/0/e/e0ed3ffa188a3630c9b3bee8926c24f4690ee573.png)
.
By cyclic exchange of elements we obtain the expressions![$\frac{[PZX]}{[ABC]}=\frac{{\overline{PB_1}} \cdot \overline{PB_2}}{4R^2} \ , \ \frac{[PXY]}{[ABC]}=\frac{{\overline{PC_1}} \cdot \overline{PC_2}}{4R^2} \ \Longrightarrow$](//latex.artofproblemsolving.com/1/0/d/10db47026362383b061d94e4883dfc34d8bbb906.png)
![$\frac{\overline{PA_1} \cdot \overline{PA_2}+\overline{PB_1} \cdot \overline{PB_2}+\overline{PC_1} \cdot \overline{PC_2}}{4R^2}=\frac{[PYZ]+[PZX]+[PXY]}{[ABC]}$](//latex.artofproblemsolving.com/4/b/b/4bbcf1ea8d2e0d0c4700d93fbd36730c76bea192.png)
. Therefore, 
![$4R^2 \cdot \frac{[XYZ]}{[ABC]}=4R^2 \cdot \frac{R^2-PO^2}{4R^2}$](//latex.artofproblemsolving.com/3/0/a/30a2a8004fbe04bc37ad6cb8b9c4ed1553e58931.png)
. In conclusion, 
.
Lemma (well-known). The power
of the point 
w.r.t. the circumcircle of is given by
, where 
- normalized barycentrical coordinates of w.r.t. 
.
Lemma. Let
be a circle and consider 
so that 
. Then 
.
Proof.
by the Stewart's theorem.
Proof. Apply lemma to
.
Remark. The proposed identity is truly generally, i.e. for any point from the plane of we"ll have 
Same generalization with other notations and other proof.
Let
be a point in the plane of with circumcircle . Consider the points 
, 
, 
so that
, 
, 
and 
. Then 
.
Proof. I"use barycentrical coordinates, i.e. I"ll consider that the point has the normalized barycentrical coordinates , where 
.
Denote
. Since 
obtain that 
. Since 
obtain that
and from 
obtain that 
and 
.
Therefore,
. Obtain analogously 
and 
.
In conclusion,
. Study some particular case, for exemple 
or 
.
be an acute triangle with the circumcircle 
. Parallel through 
to 
cuts 
, 
at 
, 
respectively and define similarly the points
, 
and 
, 
. Then 
.Proof. Sinus' theorem in
because 
. Generalization. Let
be an interior point of 
with circumcircle . Consider the points 
, 
and
so that 
, 
, 
and 
. Then 
.Proof. Let
be the orthogonal projections of onto , 
, . Denote the circumradius 
of 
. By Euler's theorem for and its pedal triangle
w.r.t. we get ![$\frac{[PYZ]}{[AA_1A_2]}=\frac{{\overline{PA_1}} \cdot \overline{PA_2}}{4{R_1}^2}$](http://latex.artofproblemsolving.com/4/5/e/45e5b48f0a3a94c9cac8898d966283f625d5e75e.png)
. But 
gives ![$\frac{[ABC]}{[A_1B_1C_1]}=\frac{R^2}{{R_1}^2}$](http://latex.artofproblemsolving.com/7/6/4/764e93bf51959ccbf8fcad571aa06ab4ea3eee8e.png)
![$\Longrightarrow \ \frac{[PYZ]}{[ABC]}=\frac{{\overline{PA_1}} \cdot \overline{PA_2}}{4R^2}$](http://latex.artofproblemsolving.com/e/0/e/e0ed3ffa188a3630c9b3bee8926c24f4690ee573.png)
.By cyclic exchange of elements we obtain the expressions
![$\frac{[PZX]}{[ABC]}=\frac{{\overline{PB_1}} \cdot \overline{PB_2}}{4R^2} \ , \ \frac{[PXY]}{[ABC]}=\frac{{\overline{PC_1}} \cdot \overline{PC_2}}{4R^2} \ \Longrightarrow$](http://latex.artofproblemsolving.com/1/0/d/10db47026362383b061d94e4883dfc34d8bbb906.png)
![$\frac{\overline{PA_1} \cdot \overline{PA_2}+\overline{PB_1} \cdot \overline{PB_2}+\overline{PC_1} \cdot \overline{PC_2}}{4R^2}=\frac{[PYZ]+[PZX]+[PXY]}{[ABC]}$](http://latex.artofproblemsolving.com/4/b/b/4bbcf1ea8d2e0d0c4700d93fbd36730c76bea192.png)
. Therefore, 
![$4R^2 \cdot \frac{[XYZ]}{[ABC]}=4R^2 \cdot \frac{R^2-PO^2}{4R^2}$](http://latex.artofproblemsolving.com/3/0/a/30a2a8004fbe04bc37ad6cb8b9c4ed1553e58931.png)
. In conclusion, 
.Lemma (well-known). The power
of the point 
w.r.t. the circumcircle of is given by
, where 
- normalized barycentrical coordinates of w.r.t. 
.Lemma. Let
be a circle and consider 
so that 
. Then 
.Proof.
by the Stewart's theorem.Proof. Apply lemma to
.Remark. The proposed identity is truly generally, i.e. for any point
from the plane of we"ll have 
Same generalization with other notations and other proof.
Let
be a point in the plane of with circumcircle . Consider the points 
, 
, 
so that
, 
, 
and 
. Then 
.Proof. I"use barycentrical coordinates, i.e. I"ll consider that the point
has the normalized barycentrical coordinates , where 
.Denote
. Since 
obtain that 
. Since 
obtain that
and from 
obtain that 
and 
.Therefore,
. Obtain analogously 
and 
.In conclusion,
. Study some particular case, for exemple 
or 
.This post has been edited 48 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:16 AM
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