Difference between revisions of "2001 AIME I Problems/Problem 5"

m
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
{{solution}}
+
Solving for y in terms of x gives <math>y=\sqrt{4-x^2}/2</math>, so the two other points of the triangle are <math>(x,\sqrt{4-x^2}/2)</math> and <math>(-x,\sqrt{4-x^2}/2)</math>, which are a distance of 2x apart. Thus 2x equals the distance between <math>(x,\sqrt{4-x^2}/2)</math> and (0,1), so by the distance formula we have <math>2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}</math>. Squaring both sides, letting <math>x^2=n</math>, and simplifying gives <math>n=192/169</math>, so <math>2x=\sqrt{768/169}</math> and the answer is 937.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2001|n=I|num-b=4|num-a=6}}

Revision as of 19:57, 17 March 2008

Problem

An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the length of each side is $\sqrt{\frac mn}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Solving for y in terms of x gives $y=\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$, which are a distance of 2x apart. Thus 2x equals the distance between $(x,\sqrt{4-x^2}/2)$ and (0,1), so by the distance formula we have $2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}$. Squaring both sides, letting $x^2=n$, and simplifying gives $n=192/169$, so $2x=\sqrt{768/169}$ and the answer is 937.

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS