Difference between revisions of "Viviani's theorem"

Revision as of 20:14, 16 May 2016

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. $[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.88, xmax = 30.12, ymin = -5.5, ymax = 11.; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((3.22,-0.78)--(14.52,-0.7)--(8.800717967697247,9.046087062764157)--cycle, zzttqq); draw((10.222698397686688,-0.3061486732450197)--(9.798444960927041,-0.30915223739907016)--(9.801448525081092,-0.7334056741587179)--(10.22570196184074,-0.7304021100046675)--cycle, qqwuqq); draw((6.871598765286669,4.790362710570685)--(7.08112432080719,5.15927874652442)--(6.712208284853454,5.368804302044939)--(6.502682729332935,4.9998882660912045)--cycle, qqwuqq); draw((11.476061820875268,3.6488294940564145)--(11.690789702114396,3.28291702225673)--(12.05670217391408,3.4976449034958583)--(11.841974292674951,3.8635573752955428)--cycle, qqwuqq); /* draw figures */ draw((3.22,-0.78)--(14.52,-0.7), zzttqq); draw((14.52,-0.7)--(8.800717967697247,9.046087062764157), zzttqq); draw((8.800717967697247,9.046087062764157)--(3.22,-0.78), zzttqq); draw((10.2,2.9)--(10.22570196184074,-0.7304021100046675)); draw((10.2,2.9)--(6.502682729332935,4.9998882660912045)); draw((10.2,2.9)--(11.841974292674951,3.8635573752955428)); label("$l+m+n = h$",(7.5043433217971725,-1.565215213000298),SE*labelscalefactor); /* dots and labels */ dot((3.22,-0.78),dotstyle); label("$A$", (3.3,-0.58), NW * labelscalefactor); dot((14.52,-0.7),dotstyle); label("$B$", (14.6,-0.5), NE * labelscalefactor); dot((8.800717967697247,9.046087062764157),dotstyle); label("$C$", (8.88,9.24), N * labelscalefactor); dot((10.2,2.9),dotstyle); label("$P$", (10.28,3.1), NE * labelscalefactor); dot((6.502682729332935,4.9998882660912045),linewidth(3.pt) + dotstyle); label("$B'$", (6.58,5.12), NW * labelscalefactor); dot((11.841974292674951,3.8635573752955428),linewidth(3.pt) + dotstyle); label("$A'$", (11.92,3.98), NE * labelscalefactor); dot((10.22570196184074,-0.7304021100046675),linewidth(3.pt) + dotstyle); label("$C'$", (10.3,-0.62), NE * labelscalefactor); label("$l$", (9.88,1.1), NE * labelscalefactor); label("$m$", (8.5,4.24), NE * labelscalefactor); label("$n$", (11.18,3.12), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$ 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 $s=AB=BC=AC$ . 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.

Page

Toolbox

Search

Difference between revisions of "Viviani's theorem"

Revision as of 20:14, 16 May 2016

Proof

Problem

@@ Line 5: / Line 5: @@
 Let <math>\triangle ABC</math> be an equilateral triangle and <math>P</math> be a point inside the triangle.
 <asy>
-pair A,B,C,P,X,Y,Z;
+ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
-real s=12*sqrt(3);
+import graph; size(8.cm);
-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);
+real labelscalefactor = 0.5; /* changes label-to-point distance */
-draw(A--B--C--cycle); draw(P--Z); draw(P--Y); draw(P--X); draw(P--A); draw(P--B); draw(P--C);
+pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
-draw(rightanglemark(P,X,B,25)); draw(rightanglemark(P,Z,C,25)); draw(rightanglemark(P,Y,A,25));
+pen dotstyle = black; /* point style */
-label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$P$",P,SE);
+real xmin = -3.88, xmax = 30.12, ymin = -5.5, ymax = 11.;  /* image dimensions */
-label("$z$",P--Z,W); label("$y$",P--X,S); label("$x$",P--Y,NE);</asy>
+pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.);
+draw((3.22,-0.78)--(14.52,-0.7)--(8.800717967697247,9.046087062764157)--cycle, zzttqq);
+draw((10.222698397686688,-0.3061486732450197)--(9.798444960927041,-0.30915223739907016)--(9.801448525081092,-0.7334056741587179)--(10.22570196184074,-0.7304021100046675)--cycle, qqwuqq);
+draw((6.871598765286669,4.790362710570685)--(7.08112432080719,5.15927874652442)--(6.712208284853454,5.368804302044939)--(6.502682729332935,4.9998882660912045)--cycle, qqwuqq);
+draw((11.476061820875268,3.6488294940564145)--(11.690789702114396,3.28291702225673)--(12.05670217391408,3.4976449034958583)--(11.841974292674951,3.8635573752955428)--cycle, qqwuqq);
+ /* draw figures */
+draw((3.22,-0.78)--(14.52,-0.7), zzttqq);
+draw((14.52,-0.7)--(8.800717967697247,9.046087062764157), zzttqq);
+draw((8.800717967697247,9.046087062764157)--(3.22,-0.78), zzttqq);
+draw((10.2,2.9)--(10.22570196184074,-0.7304021100046675));
+draw((10.2,2.9)--(6.502682729332935,4.9998882660912045));
+draw((10.2,2.9)--(11.841974292674951,3.8635573752955428));
+label("$l+m+n = h$",(7.5043433217971725,-1.565215213000298),SE*labelscalefactor);
+ /* dots and labels */
+dot((3.22,-0.78),dotstyle);
+label("$A$", (3.3,-0.58), NW * labelscalefactor);
+dot((14.52,-0.7),dotstyle);
+label("$B$", (14.6,-0.5), NE * labelscalefactor);
+dot((8.800717967697247,9.046087062764157),dotstyle);
+label("$C$", (8.88,9.24), N * labelscalefactor);
+dot((10.2,2.9),dotstyle);
+label("$P$", (10.28,3.1), NE * labelscalefactor);
+dot((6.502682729332935,4.9998882660912045),linewidth(3.pt) + dotstyle);
+label("$B'$", (6.58,5.12), NW * labelscalefactor);
+dot((11.841974292674951,3.8635573752955428),linewidth(3.pt) + dotstyle);
+label("$A'$", (11.92,3.98), NE * labelscalefactor);
+dot((10.22570196184074,-0.7304021100046675),linewidth(3.pt) + dotstyle);
+label("$C'$", (10.3,-0.62), NE * labelscalefactor);
+label("$l$", (9.88,1.1), NE * labelscalefactor);
+label("$m$", (8.5,4.24), NE * labelscalefactor);
+label("$n$", (11.18,3.12), NE * labelscalefactor);
+clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
+ /* end of picture */</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>. 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.