Difference between revisions of "Viviani's theorem"
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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> | ||
− | + | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | |
− | real | + | import graph; size(8.cm); |
− | + | real labelscalefactor = 0.5; /* changes label-to-point distance */ | |
− | draw( | + | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ |
− | draw( | + | pen dotstyle = black; /* point style */ |
− | label("$A$", | + | real xmin = -3.88, xmax = 30.12, ymin = -5.5, ymax = 11.; /* image dimensions */ |
− | label("$ | + | 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. | 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. | ||
Revision as of 19:14, 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 be an equilateral triangle and be a point inside the triangle. We label the altitudes from to each of sides , and , and respectively. Since is equilateral, we can say that . Therefore, , and . Since the area of a triangle is the product of its base and altitude, we also have . However, the area of can also be expressed as . 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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