https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathlover3737&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T22:35:24ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Area_of_an_equilateral_triangle&diff=95274Area of an equilateral triangle2018-06-18T08:06:42Z<p>Mathlover3737: </p>
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<div>The area of an equilateral triangle is <math>\frac{s^2\sqrt{3}}{4}</math>, where <math>s</math> is the sidelength of the triangle.<br />
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== Proof ==<br />
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''Method 1:'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is <math>\frac{s}{2}</math>.<br />
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Using the [[Pythagorean theorem]], we get <math>s^2=h^2+\frac{s^2}{4}</math>, where <math>h</math> is the height of the triangle. Solving, <math>h=\frac{s\sqrt{3}}{2}</math>. (note we could use 30-60-90 right triangles.)<br />
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We use the formula for the area of a triangle, <math>\frac{1}{2}bh</math> (note that <math>s</math> is the length of a base), so the area is <cmath>\frac{1}{2}(s)\left(\frac{s\sqrt{3}}{2}\right) = \boxed{\frac{s^2\sqrt{3}}{4}}</cmath><br />
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''Method 2:'' '''(warning: uses trig.)''' The area of a triangle is <math>\frac{ab\sin{C}}{2}</math>. Plugging in <math>a=b=s</math> and <math>C=\frac{\pi}{3}</math> (the angle at each vertex, in radians), we get the area to be <math>\frac{s^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}</math></div>Mathlover3737https://artofproblemsolving.com/wiki/index.php?title=Area_of_an_equilateral_triangle&diff=95273Area of an equilateral triangle2018-06-18T08:06:05Z<p>Mathlover3737: Add some details</p>
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<div>The area of an equilateral triangle is <math>\frac{s^2\sqrt{3}}{4}</math>, where <math>s</math> is the sidelength of the triangle.<br />
<br />
<br />
== Proof ==<br />
<br />
''Method 1:'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is <math>\frac{s}{2}</math>.<br />
<br />
Using the [[Pythagorean theorem]], we get <math>s^2=h^2+\frac{s^2}{4}</math>, where <math>h</math> is the height of the triangle. Solving, <math>h=\frac{s\sqrt{3}}{2}</math>. (note we could use 30-60-90 right triangles.)<br />
<br />
We use the formula for the area of a triangle, <math>\frac{1}{2}bh</math> (note <math>s</math> is the length of a base), so the area is <cmath>\frac{1}{2}(s)\left(\frac{s\sqrt{3}}{2}\right) \boxed{\frac{s^2\sqrt{3}}{4}}</cmath><br />
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''Method 2:'' '''(warning: uses trig.)''' The area of a triangle is <math>\frac{ab\sin{C}}{2}</math>. Plugging in <math>a=b=s</math> and <math>C=\frac{\pi}{3}</math> (the angle at each vertex, in radians), we get the area to be <math>\frac{s^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}</math></div>Mathlover3737https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_5&diff=529352013 AIME II Problems/Problem 52013-06-04T03:46:30Z<p>Mathlover3737: /* Solution */</p>
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<div>==Problem 5==<br />
In equilateral <math>\triangle ABC</math> let points <math>D</math> and <math>E</math> trisect <math>\overline{BC}</math>. Then <math>\sin(\angle DAE)</math> can be expressed in the form <math>\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is an integer that is not divisible by the square of any prime. Find <math>a+b+c</math>.<br />
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== Solution ==<br />
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<asy><br />
pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0);<br />
pair M = (1, 0);<br />
pair D = (2/3, 0), E = (4/3, 0);<br />
draw(A--B--C--cycle);<br />
label("$A$", A, N);<br />
label("$B$", B, SW);<br />
label("$C$", C, SE);<br />
label("$D$", D, S);<br />
label("$E$", E, S);<br />
label("$M$", M, S);<br />
draw(A--D);<br />
draw(A--E);<br />
draw(A--M);</asy><br />
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Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier. <br />
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Let <math>M</math> be the midpoint of <math>\overline{DE}</math>. Then <math>\Delta MCA</math> is a 30-60-90 triangle with <math>MC = \dfrac{3}{2}</math>, <math>AC = 3</math> and <math>AM = \dfrac{3\sqrt{3}}{2}</math>. Since the triangle <math>\Delta AME</math> is right, then we can find the length of <math>\overline{AE}</math> by pythagorean theorem, <math>AE = \sqrt{7}</math>. Therefore, since <math>\Delta AME</math> is a right triangle, we can easily find <math>\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}</math> and <math>\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}</math>. So we can use the double angle formula for sine, <math>\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>.<br />
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==See Also==<br />
{{AIME box|year=2013|n=II|num-b=4|num-a=6}}</div>Mathlover3737https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&diff=504322009 AIME I Problems/Problem 42013-01-02T04:11:55Z<p>Mathlover3737: /* Solution 1 */</p>
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<div>== Problem 4 ==<br />
In parallelogram <math>ABCD</math>, point <math>M</math> is on <math>\overline{AB}</math> so that <math>\frac {AM}{AB} = \frac {17}{1000}</math> and point <math>N</math> is on <math>\overline{AD}</math> so that <math>\frac {AN}{AD} = \frac {17}{2009}</math>. Let <math>P</math> be the point of intersection of <math>\overline{AC}</math> and <math>\overline{MN}</math>. Find <math>\frac {AC}{AP}</math>.<br />
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==Solution==<br />
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===Solution 1===<br />
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One of the ways to solve this problem is to make this parallelogram a straight line.<br />
So the whole length of the line is <math>APC</math>(<math>AMC</math> or <math>ANC</math>), and <math>ABC</math> is <math>1000x+2009x=3009x.</math><br />
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<math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x.</math><br />
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So the answer is <math>3009x/17x = \boxed{177}</math><br />
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===Solution 2===<br />
Draw a diagram with all the given points and lines involved. Construct parallel lines <math>\overline{DF_2F_1}</math> and <math>\overline{BB_1B_2}</math> to <math>\overline{MN}</math>, where for the lines the endpoints are on <math>\overline{AM}</math> and <math>\overline{AN}</math>, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral <math>BB_2DF_1</math> <math>\overline{E_1E_2E_3}</math> where the points are in order from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2009}{17}MN</math>. It is not difficult to see that <math>E_2</math> is the center of quadrilateral <math>ABCD</math> and thus the midpoint of <math>\overline{AC}</math> as well as the midpoint of <math>\overline{B_1}{F_2}</math> (all of this is easily proven with symmetry). From more triangle similarity, <math>E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP</math><br />
<math>= \boxed{177}AP</math>.<br />
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===Solution 3===<br />
Using vectors, note that <math>\overrightarrow{AM}=\frac{17}{1000}\overrightarrow{AB}</math> and <math>\overrightarrow{AN}=\frac{17}{2009}\overrightarrow{AD}</math>. Note that <math>\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}</math> for some positive x and y, but at the same time is a scalar multiple of <math>\overrightarrow{AB}+\overrightarrow{AD}</math>. So, writing the equation <math>\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}</math> in terms of <math>\overrightarrow{AB}</math> and <math>\overrightarrow{AD}</math>, we have <math>\overrightarrow{AP}=\frac{\frac{17x}{1000}\overrightarrow{AB}+\frac{17y}{2009}\overrightarrow{AD}}{x+y}</math>. But the coefficients of the two vectors must be equal because, as already stated, <math>\overrightarrow{AP}</math> is a scalar multiple of <math>\overrightarrow{AB}+\overrightarrow{AD}</math>. We then see that <math>\frac{x}{x+y}=\frac{1000}{3009}</math> and <math>\frac{y}{x+y}=\frac{2009}{3009}</math>. Finally, we have <math>\overrightarrow{AP}=\frac{17}{3009}(\overrightarrow{AB}+\overrightarrow{AD})</math> and, simplifying, <math>\overrightarrow{AB}+\overrightarrow{AD}}=177\overrightarrow{AP}</math> and the desired quantity is <math>177</math>.<br />
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== See also ==<br />
{{AIME box|year=2009|n=I|num-b=3|num-a=5}}</div>Mathlover3737