80. Three concurrent perpendiculars (the Carnot's lemma).
by Virgil Nicula, Aug 5, 2010, 2:27 PM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=360004
P1. Let
be a non-right triangle with the orthocenter 
and let 
be its median triangle, where 
, 
, 
. Denote
, 
, 
. Prove that the perpendiculars from 
, 
, 
to 
, 
, 
are concurrent.
Proof. Denote
, 
, 
and 
, 
, 
. Observe that 
and
. Therefore, 
.
Using Carnot's lemma results the conclusion, i.e. the perpendiculars from , , to , , are concurrent.
P2 (Miguel Ochoa Sanchez). Let an acute
and three interior points 
so that 
Denote
Prove that ![$\sum x^2-\sum a^2=4\cdot \sqrt{[MNP]\cdot [ABC]}\ .$](//latex.artofproblemsolving.com/3/d/5/3d5990ff545a433d13a5a6acfb7c241b6da36d82.png)
Proof.Let
Thus, 
i.e. ![$\frac {MN}{BC}=\frac {NP}{CA}=\frac {PM}{AB}\equiv k\equiv \frac {h_p}{h_a}=\frac {h_m}{h_b}=\frac {h_n}{h_c}=\sqrt{\frac {[MNP]}{[ABC]}}$](//latex.artofproblemsolving.com/0/b/2/0b2acf7ca513dc928eee5056bd6737e0f9175f56.png)
where 
is the ratio
of similarity. So![$\left\{\begin{array}{c}
2\cdot [MNP]=MN\cdot h_p=NP\cdot h_m=PM\cdot h_n\\\\
2\cdot [ABC]=BC\cdot h_a=CA\cdot h_b=AB\cdot h_c\end{array}\right\|$](//latex.artofproblemsolving.com/8/c/1/8c18c07141539dde15dd6e434065b0ca99463dd8.png)
Apply Carnot's theorem to 
Therefore,![$\frac {[MNP]}{[ABC]}=\frac {MN\cdot h_p}{BC\cdot h_a}=k^2\ ,$](//latex.artofproblemsolving.com/5/0/e/50ef960d178adbd9b8d13c81e395a47990c2bf79.png)
i.e. ![$\boxed{[MNP]=k^2\cdot [ABC]}\ (1)\ .$](//latex.artofproblemsolving.com/0/a/d/0ad9d4cd85726576278ca813fe5d13d1055fee02.png)
Thus, ![$AB\cdot h_n=\frac {2[ABC]}{h_c}\cdot h_n=$](//latex.artofproblemsolving.com/f/1/4/f142dea09de2e261c159a4a4872b2e65a2b539b9.png)
![$2[ABC]\cdot k=$](//latex.artofproblemsolving.com/f/4/3/f438fdaa7771c9012bc54707f077092f3b24d474.png)
![$2[ABC]\cdot\sqrt{\frac{[MNP]}{[ABC]}}=$](//latex.artofproblemsolving.com/2/c/1/2c1519d51a2dd918d1bcbb734bd230f12ccc6df5.png)
![$2\cdot\sqrt{[MNP]\cdot[ABC]}\ ,$](//latex.artofproblemsolving.com/c/4/0/c4052328d352b9f65cb9f58594ad34a8f8965b4e.png)
i.e. (analogously)![$AB\cdot h_n=BC\cdot h_p=CA\cdot h_m=2\cdot\sqrt{[MNP]\cdot[ABC]}\ .$](//latex.artofproblemsolving.com/8/f/d/8fd95d2e31f5af69a29bb3f1084668606b02e35e.png)
In conclusion, ![$\sum x^2-\sum a^2=\frac 23\cdot\sum AB\cdot h_n=4\cdot\sqrt{[MNP]\cdot[ABC]}$](//latex.artofproblemsolving.com/c/4/7/c470b4004d0ed18f54cc22c0917b0b387a8f00e3.png)
(Baris Altay).
P1. Let
be a non-right triangle with the orthocenter 
and let 
be its median triangle, where 
, 
, 
. Denote
, 
, 
. Prove that the perpendiculars from 
, 
, 
to 
, 
, 
are concurrent.Proof. Denote
, 
, 
and 
, 
, 
. Observe that 
and
. Therefore, 
.Using Carnot's lemma results the conclusion, i.e. the perpendiculars from
, , to , , are concurrent.P2 (Miguel Ochoa Sanchez). Let an acute
and three interior points 
so that 
Denote
Prove that ![$\sum x^2-\sum a^2=4\cdot \sqrt{[MNP]\cdot [ABC]}\ .$](http://latex.artofproblemsolving.com/3/d/5/3d5990ff545a433d13a5a6acfb7c241b6da36d82.png)
Proof.Let
Thus, 
i.e. ![$\frac {MN}{BC}=\frac {NP}{CA}=\frac {PM}{AB}\equiv k\equiv \frac {h_p}{h_a}=\frac {h_m}{h_b}=\frac {h_n}{h_c}=\sqrt{\frac {[MNP]}{[ABC]}}$](http://latex.artofproblemsolving.com/0/b/2/0b2acf7ca513dc928eee5056bd6737e0f9175f56.png)
where 
is the ratioof similarity. So
![$\left\{\begin{array}{c}
2\cdot [MNP]=MN\cdot h_p=NP\cdot h_m=PM\cdot h_n\\\\
2\cdot [ABC]=BC\cdot h_a=CA\cdot h_b=AB\cdot h_c\end{array}\right\|$](http://latex.artofproblemsolving.com/8/c/1/8c18c07141539dde15dd6e434065b0ca99463dd8.png)
Apply Carnot's theorem to 
Therefore,
![$\frac {[MNP]}{[ABC]}=\frac {MN\cdot h_p}{BC\cdot h_a}=k^2\ ,$](http://latex.artofproblemsolving.com/5/0/e/50ef960d178adbd9b8d13c81e395a47990c2bf79.png)
i.e. ![$\boxed{[MNP]=k^2\cdot [ABC]}\ (1)\ .$](http://latex.artofproblemsolving.com/0/a/d/0ad9d4cd85726576278ca813fe5d13d1055fee02.png)
Thus, ![$AB\cdot h_n=\frac {2[ABC]}{h_c}\cdot h_n=$](http://latex.artofproblemsolving.com/f/1/4/f142dea09de2e261c159a4a4872b2e65a2b539b9.png)
![$2[ABC]\cdot k=$](http://latex.artofproblemsolving.com/f/4/3/f438fdaa7771c9012bc54707f077092f3b24d474.png)
![$2[ABC]\cdot\sqrt{\frac{[MNP]}{[ABC]}}=$](http://latex.artofproblemsolving.com/2/c/1/2c1519d51a2dd918d1bcbb734bd230f12ccc6df5.png)
![$2\cdot\sqrt{[MNP]\cdot[ABC]}\ ,$](http://latex.artofproblemsolving.com/c/4/0/c4052328d352b9f65cb9f58594ad34a8f8965b4e.png)
i.e. (analogously)
![$AB\cdot h_n=BC\cdot h_p=CA\cdot h_m=2\cdot\sqrt{[MNP]\cdot[ABC]}\ .$](http://latex.artofproblemsolving.com/8/f/d/8fd95d2e31f5af69a29bb3f1084668606b02e35e.png)
In conclusion, ![$\sum x^2-\sum a^2=\frac 23\cdot\sum AB\cdot h_n=4\cdot\sqrt{[MNP]\cdot[ABC]}$](http://latex.artofproblemsolving.com/c/4/7/c470b4004d0ed18f54cc22c0917b0b387a8f00e3.png)
(Baris Altay).
This post has been edited 59 times. Last edited by Virgil Nicula, Aug 26, 2016, 1:25 AM
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