Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Viviani's theorem"
(stating this article is a stub)
Line 17: Line 17:
  
 
[http://www.artofproblemsolving.com/Forum/viewtopic.php?type=alcumus&id=41565]
 
[http://www.artofproblemsolving.com/Forum/viewtopic.php?type=alcumus&id=41565]
 +
 +
{{stub}}

Revision as of 18:39, 12 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

[1]

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Viviani%27s_theorem&oldid=49499"