Difference between revisions of "2001 AIME II Problems/Problem 6"
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== Problem == | == Problem == | ||
− | [[Square]] <math>ABCD</math> is inscribed in a [[circle]]. Square <math>EFGH</math> has vertices <math>E</math> and <math>F</math> on <math>\overline{CD}</math> and vertices <math>G</math> and <math>H</math> on the circle. | + | [[Square]] <math>ABCD</math> is inscribed in a [[circle]]. Square <math>EFGH</math> has vertices <math>E</math> and <math>F</math> on <math>\overline{CD}</math> and vertices <math>G</math> and <math>H</math> on the circle. If the area of square <math>ABCD</math> is <math>1</math>, then the area of square <math>EFGH</math> can be expressed as <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers and <math>m < n</math>. Find <math>10n + m</math>. |
− | == Solution == | + | == Solution 1(Pythagorean Theorem)== |
Let <math>O</math> be the center of the circle, and <math>2a</math> be the side length of <math>ABCD</math>, <math>2b</math> be the side length of <math>EFGH</math>. By the [[Pythagorean Theorem]], the radius of <math>\odot O = OC = a\sqrt{2}</math>. | Let <math>O</math> be the center of the circle, and <math>2a</math> be the side length of <math>ABCD</math>, <math>2b</math> be the side length of <math>EFGH</math>. By the [[Pythagorean Theorem]], the radius of <math>\odot O = OC = a\sqrt{2}</math>. | ||
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<cmath>0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2</cmath> | <cmath>0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2</cmath> | ||
This is a quadratic in <math>\frac{b}{a}</math>, and solving it gives <math>\frac{b}{a} = \frac{1}{5},-1</math>. The negative solution is extraneous, and so the ratio of the areas is <math>\left(\frac{1}{5}\right)^2 = \frac{1}{25}</math> and the answer is <math>10\cdot 25 + 1 = \boxed{251}</math>. | This is a quadratic in <math>\frac{b}{a}</math>, and solving it gives <math>\frac{b}{a} = \frac{1}{5},-1</math>. The negative solution is extraneous, and so the ratio of the areas is <math>\left(\frac{1}{5}\right)^2 = \frac{1}{25}</math> and the answer is <math>10\cdot 25 + 1 = \boxed{251}</math>. | ||
+ | |||
+ | Remark: The division by <math>a^2</math> is equivalent to simply setting the original area of square <math>ABCD</math> to 1. | ||
+ | |||
+ | == Solution 2 (Coordinates) == | ||
+ | Let point <math>A</math> be the top-left corner of square <math>ABCD</math> and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let <math>D</math> have coordinates <math>(0,0)</math> and the side length of square <math>ABCD</math> be <math>a</math>. Let <math>DF</math> = <math>b</math> and diameter <math>HI</math> go through <math>J</math> the midpoint of <math>EF</math>. Since a diameter always bisects a chord perpendicular to it, <math>DJ</math> = <math>JC</math> and since <math>F</math> and <math>E</math> must be symmetric around the diameter, <math>FJ = JE</math> and it follows that <math>DF = EC = b.</math> Hence <math>FE</math> the side of square <math>EFGH</math> has length <math>a - 2b</math>. <math>F</math> has coordinates <math>(b,0)</math> and <math>G</math> has coordinates <math>(b, 2b - a).</math> We know that point <math>G</math> must be on the circle <math>O</math> - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square <math>(a/2, a/2)</math> and has radius <math>a * </math><math>\sqrt{2} / 2</math>, half the diagonal of the square, | ||
+ | <math>(x - a/2)^2 + (y - a/2)^2 = 1/2a^2</math> follows as the circle equation. Then substituting coordinates of <math>G</math> into the equation, <math>(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2</math>. Simplifying and factoring, we get <math>2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.</math> Since <math>a = b</math> would imply <math>m = n</math>, and <math>m < n</math> in the problem, we must use the other factor. We get <math>b = 2/5a</math>, meaning the ratio of areas <math>((a-2b)/a)^2</math> = <math>(1/5)^2</math> = <math>1/25</math> = <math>m/n.</math> Then <math>10n + m = 25 * 10 + 1 = \boxed{251}</math>. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:31, 6 December 2023
Problem
Square is inscribed in a circle. Square
has vertices
and
on
and vertices
and
on the circle. If the area of square
is
, then the area of square
can be expressed as
where
and
are relatively prime positive integers and
. Find
.
Solution 1(Pythagorean Theorem)
Let be the center of the circle, and
be the side length of
,
be the side length of
. By the Pythagorean Theorem, the radius of
.
![[asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); [/asy]](http://latex.artofproblemsolving.com/a/1/4/a14d609f4250e7310e06648a5d575443252da2bb.png)
Now consider right triangle , where
is the midpoint of
. Then, by the Pythagorean Theorem,
Thus (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so
, and the answer is
.
Another way to proceed from is to note that
is the quantity we need; thus, we divide by
to get
This is a quadratic in
, and solving it gives
. The negative solution is extraneous, and so the ratio of the areas is
and the answer is
.
Remark: The division by is equivalent to simply setting the original area of square
to 1.
Solution 2 (Coordinates)
Let point be the top-left corner of square
and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let
have coordinates
and the side length of square
be
. Let
=
and diameter
go through
the midpoint of
. Since a diameter always bisects a chord perpendicular to it,
=
and since
and
must be symmetric around the diameter,
and it follows that
Hence
the side of square
has length
.
has coordinates
and
has coordinates
We know that point
must be on the circle
- hence it must satisfy the circle equation. Since the center of the circle is at the center of the square
and has radius
, half the diagonal of the square,
follows as the circle equation. Then substituting coordinates of
into the equation,
. Simplifying and factoring, we get
Since
would imply
, and
in the problem, we must use the other factor. We get
, meaning the ratio of areas
=
=
=
Then
.
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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