Difference between revisions of "2017 AIME I Problems/Problem 15"
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==Solution== | ==Solution== | ||
| + | Lemma. If <math>x,y</math> satisfy <math>px+qy=1</math>, then the minimal value of <math>\sqrt{x^2+y^2}</math> is <math>\frac{1}{\sqrt{p^2+q^2}}</math>. | ||
| + | |||
| + | Proof. Recall that the distance between the point <math>(x_0,y_0)</math> and the line <math>px+qy+r = 0</math> is given by <math>\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}</math>. In particular, the distance between the origin and any point <math>(x,y)</math> on the line <math>px+qy=1</math> is at least <math>\frac{1}{\sqrt{p^2+q^2}}</math>. | ||
| + | |||
| + | --- | ||
| + | |||
| + | Let the vertices of the right triangle be <math>(0,0),(5,0),(0,2\sqrt{3}),</math> and let <math>(a,0),(0,b)</math> be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is <math>\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)</math>. This point must lie on the hypotenuse <math>\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1</math>, i.e. <math>a,b</math> must satisfy | ||
| + | <cmath> \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,</cmath> | ||
| + | which can be simplified to | ||
| + | <cmath>\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.</cmath> | ||
| + | |||
| + | By the lemma, the minimal value of <math>\sqrt{a^2+b^2}</math> is | ||
| + | <cmath>\frac{1}{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2} = \frac{10\sqrt{3}}{\sqrt{67}},</cmath> | ||
| + | so the minimal area of the equilateral triangle is | ||
| + | <cmath> \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},</cmath> | ||
| + | and hence the answer is <math>75+3+67=\boxed{145}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:31, 9 March 2017
Problem 15
The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths 
and 
as shown, is 
where 
and 
are positive integers, 
and 
are relatively prime, and is not divisible by the square of any prime. Find 
Solution
Lemma. If 
satisfy 
, then the minimal value of 
is 
.
Proof. Recall that the distance between the point 
and the line 
is given by 
. In particular, the distance between the origin and any point 
on the line is at least .
---
Let the vertices of the right triangle be 
and let 
be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is 
. This point must lie on the hypotenuse 
, i.e. 
must satisfy
![\[\frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,\]](http://latex.artofproblemsolving.com/f/e/6/fe6a7386eb66a3b42663a6cc53280e0641ec470d.png)
which can be simplified to
![\[\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.\]](http://latex.artofproblemsolving.com/d/b/9/db911e22c50d482d2ed62526f19a435bfb26ede0.png)
By the lemma, the minimal value of 
is
![\[\frac{1}{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2} = \frac{10\sqrt{3}}{\sqrt{67}},\]](http://latex.artofproblemsolving.com/7/0/5/705d39b6b44e32a2c92cba59bc16226d41bcaea5.png)
so the minimal area of the equilateral triangle is
![\[\frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},\]](http://latex.artofproblemsolving.com/f/a/2/fa2164930d825d52eda5643945d73784f51576ea.png)
and hence the answer is 
.
See Also
| 2017 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
