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Difference between revisions of "Viviani's theorem"

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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.
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'''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 ==
 
== Proof ==
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Let <math>\triangle ABC</math> be an equilateral triangle and <math>P</math> be a point inside the triangle.
 
Let <math>\triangle ABC</math> be an equilateral triangle and <math>P</math> be a point inside the triangle.
 
<asy>
 
<asy>
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pair A,B,C,P,X,Y,Z;
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/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
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real s=12*sqrt(3);
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import graph; size(8.cm);  
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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);
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real labelscalefactor = 0.5; /* changes label-to-point distance */
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draw(A--B--C--cycle); draw(P--Z); draw(P--Y); draw(P--X); draw(P--A); draw(P--B); draw(P--C);
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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draw(rightanglemark(P,X,B,25)); draw(rightanglemark(P,Z,C,25)); draw(rightanglemark(P,Y,A,25));  
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pen dotstyle = black; /* point style */
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label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$P$",P,SE);
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real xmin = -3.88, xmax = 30.12, ymin = -5.5, ymax = 11.; /* image dimensions */
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label("$z$",P--Z,W); label("$y$",P--X,S); label("$x$",P--Y,NE);</asy>
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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("$x+y+z = a$",(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);  
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label("$P$", (10.28,3.1), NE * labelscalefactor);
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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("$x$", (9.88,1.1), NE * labelscalefactor);
 +
label("$z$", (8.5,4.24), NE * labelscalefactor);
 +
label("$y$", (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.
 
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.
  
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== Problem ==
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==See also==
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[[Category:Theorems]]
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[http://www.artofproblemsolving.com/Forum/viewtopic.php?type=alcumus&id=41565 Alcumus Problem]
 
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{{stub}}
 

Latest revision as of 13:13, 4 June 2021

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("$x+y+z = a$",(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("$x$", (9.88,1.1), NE * labelscalefactor); label("$z$", (8.5,4.24), NE * labelscalefactor); label("$y$", (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.

See also

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