Difference between revisions of "Viviani's theorem"

Revision as of 23:52, 19 November 2012

The Viviani's Theorem states that for an equilateral triangle, the sum of the altitudes from any point in the triangle is equal to the altitude from a vertex of the triangle to the other side.

Proof

Let $\triangle ABC$ be an equilateral triangle and $P$ be a point inside the triangle. $[asy] pair A,B,C,P,X,Y,Z; real s=12*sqrt(3); A=(0,0); C=(s,0); B=(s/2,s/2*sqrt(3)); P=(9.5,7); X= foot(P,B,C); Y=foot(P,A,B); Z=foot(P,A,C); draw(A--B--C--cycle); draw(P--Z); draw(P--Y); draw(P--X); draw(P--A); draw(P--B); draw(P--C); draw(rightanglemark(P,X,B,25)); draw(rightanglemark(P,Z,C,25)); draw(rightanglemark(P,Y,A,25)); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$P$",P,SE); label("$z$",P--Z,W); label("$y$",P--X,S); label("$x$",P--Y,NE);[/asy]$ We label the altitudes from $P$ to each of sides $\overline{AB}$ , $\overline{BC}$ and $\overline{AC}$ $x$ , $y$ and $z$ respectively. Since $\triangle ABC$ is equilateral, we can say that $s=AB=BC=AC$ WLOG. Therefore, $[ABP]=\dfrac{sx}{2}$ , $[BCP]=\dfrac{sy}{2}$ and $[ACP]=\dfrac{sz}{2}$ . Since the area of a triangle is the product of its base and altitude, we also have $[ABC]=\dfrac{as}{2}$ . However, the area of $\triangle ABC$ can also be expressed as $[ABC]=[ABP]+[BCP]+[ACP]=\dfrac{sx}{2}+\dfrac{sy}{2}+\dfrac{sz}{2}=\dfrac{s}{2}(x+y+z)$ . Therefore, $\dfrac{s}{2}(x+y+z)=\dfrac{s}{2}(a)$ , so $x+y+z=a$ , which means the sum of the altitudes from any point within the triangle is equal to the altitude from the vertex of a triangle.

Problem

Alcumus Problem

This article is a stub. Help us out by expanding it.

@@ Line 16: / Line 16: @@
 == Problem ==
-[http://www.artofproblemsolving.com/Forum/viewtopic.php?type=alcumus&id=41565]
+[http://www.artofproblemsolving.com/Forum/viewtopic.php?type=alcumus&id=41565 Alcumus Problem]
 {{stub}}

Page

Toolbox

Search

Difference between revisions of "Viviani's theorem"

Revision as of 23:52, 19 November 2012

Proof

Problem