Difference between revisions of "Viviani's theorem"
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label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$P$",P,SE); | 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> | label("$z$",P--Z,W); label("$y$",P--X,S); label("$x$",P--Y,NE);</asy> | ||
| − | We label the altitudes from <math>P</math> to each of sides <math>\overline{AB}</math>, <math>\overline{BC}</math> and <math>\overline{AC}</math> <math>x</math>, <math>y</math> and <math>z</math> respectively. Since <math>\triangle ABC</math> is equilateral, we can say that <math>s=AB=BC=AC</math> | + | We label the altitudes from <math>P</math> to each of sides <math>\overline{AB}</math>, <math>\overline{BC}</math> and <math>\overline{AC}</math> <math>x</math>, <math>y</math> and <math>z</math> respectively. Since <math>\triangle ABC</math> is equilateral, we can say that <math>s=AB=BC=AC</math>. Therefore, <math>[ABP]=\dfrac{sx}{2}</math>, <math>[BCP]=\dfrac{sy}{2}</math> and <math>[ACP]=\dfrac{sz}{2}</math>. Since the area of a triangle is the product of its base and altitude, we also have <math>[ABC]=\dfrac{as}{2}</math>. However, the area of <math>\triangle ABC</math> can also be expressed as <math>[ABC]=[ABP]+[BCP]+[ACP]=\dfrac{sx}{2}+\dfrac{sy}{2}+\dfrac{sz}{2}=\dfrac{s}{2}(x+y+z)</math>. Therefore, <math>\dfrac{s}{2}(x+y+z)=\dfrac{s}{2}(a)</math>, so <math>x+y+z=a</math>, 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 == | == Problem == | ||
Revision as of 01:50, 4 February 2013
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 
be an equilateral triangle and 
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]](http://latex.artofproblemsolving.com/d/b/3/db38f3768cd250094d664be9ece7a977dc033a7c.png)
We label the altitudes from to each of sides 
, 
and 
, 
and 
respectively. Since is equilateral, we can say that 
. Therefore, ![$[ABP]=\dfrac{sx}{2}$](http://latex.artofproblemsolving.com/3/e/1/3e1965aab6649a16f2534cfd3003f135a438096b.png)
, ![$[BCP]=\dfrac{sy}{2}$](http://latex.artofproblemsolving.com/3/8/0/380dcf10a2b74e7a65a5db42157b3efb498de67f.png)
and ![$[ACP]=\dfrac{sz}{2}$](http://latex.artofproblemsolving.com/d/8/6/d8600fc6537941ed1426e92a870f878929cf3c8a.png)
. Since the area of a triangle is the product of its base and altitude, we also have ![$[ABC]=\dfrac{as}{2}$](http://latex.artofproblemsolving.com/6/b/7/6b7488e59253a02cc677381a389700bea38ed51f.png)
. However, the area of can also be expressed as ![$[ABC]=[ABP]+[BCP]+[ACP]=\dfrac{sx}{2}+\dfrac{sy}{2}+\dfrac{sz}{2}=\dfrac{s}{2}(x+y+z)$](http://latex.artofproblemsolving.com/5/9/4/594cc7628b983dd48312f008f71de96e83c80c16.png)
. Therefore, 
, so 
, 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
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