Difference between revisions of "2023 AIME I Problems/Problem 5"
MRENTHUSIASM (talk | contribs) |
R00tsofunity (talk | contribs) (Solutions should be first-come-first-serve.) |
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(Available Soon) | (Available Soon) | ||
− | ==Solution 1 | + | ==Solution 1 (Ptolemy's Theorem)== |
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[[Ptolemy's theorem]] states that for cyclic quadrilateral <math>WXYZ</math>, <math>WX\cdot YZ + XY\cdot WZ = WY\cdot XZ</math>. | [[Ptolemy's theorem]] states that for cyclic quadrilateral <math>WXYZ</math>, <math>WX\cdot YZ + XY\cdot WZ = WY\cdot XZ</math>. | ||
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~mathboy100 | ~mathboy100 | ||
− | ==Solution | + | ==Solution 2 (Heights and Half-Angle Formula)== |
Drop a height from point <math>P</math> to line <math>\overline{AC}</math> and line <math>\overline{BC}</math>. Call these two points to be <math>X</math> and <math>Y</math>, respectively. Notice that the intersection of the diagonals of <math>\square ABCD</math> meets at a right angle at the center of the circumcircle, call this intersection point <math>O</math>. | Drop a height from point <math>P</math> to line <math>\overline{AC}</math> and line <math>\overline{BC}</math>. Call these two points to be <math>X</math> and <math>Y</math>, respectively. Notice that the intersection of the diagonals of <math>\square ABCD</math> meets at a right angle at the center of the circumcircle, call this intersection point <math>O</math>. | ||
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~Danielzh | ~Danielzh | ||
− | ==Solution | + | ==Solution 3 (Analytic Geometry)== |
Denote by <math>x</math> the half length of each side of the square. | Denote by <math>x</math> the half length of each side of the square. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution | + | ==Solution 4 (Law of Cosines)== |
WLOG, let <math>P</math> be on minor arc <math>\overarc {AB}</math>. Let <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. | WLOG, let <math>P</math> be on minor arc <math>\overarc {AB}</math>. Let <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. | ||
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~OrangeQuail9 | ~OrangeQuail9 | ||
− | ==Solution | + | ==Solution 5 (Similar Triangles)== |
Someone please help render this | Someone please help render this | ||
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~Cantalon | ~Cantalon | ||
− | ==Solution | + | ==Solution 6 (Subtended Chords)== |
First draw a diagram. | First draw a diagram. | ||
<asy> | <asy> | ||
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<cmath>2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}</cmath> | <cmath>2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}</cmath> | ||
~Voldemort101 | ~Voldemort101 | ||
+ | |||
+ | ==Solution 7 (Areas and Pythagorean Theorem)== | ||
==See also== | ==See also== |
Revision as of 18:00, 8 February 2023
Contents
Problem
Let be a point on the circle circumscribing square
that satisfies
and
Find the area of
Diagram
(Available Soon)
Solution 1 (Ptolemy's Theorem)
Ptolemy's theorem states that for cyclic quadrilateral ,
.
We may assume that is between
and
. Let
,
,
,
, and
. We have
, because
is a diameter of the circle. Similarly,
. Therefore,
. Similarly,
.
By Ptolemy's Theorem on ,
, and therefore
. By Ptolemy's on
,
, and therefore
. By squaring both equations, we obtain
Thus, , and
. Plugging these values into
, we obtain
, and
. Now, we can solve using
and
(though using
and
yields the same solution for
).
The answer is .
~mathboy100
Solution 2 (Heights and Half-Angle Formula)
Drop a height from point to line
and line
. Call these two points to be
and
, respectively. Notice that the intersection of the diagonals of
meets at a right angle at the center of the circumcircle, call this intersection point
.
Since is a rectangle,
is the distance from
to line
. We know that
by triangle area and given information. Then, notice that the measure of
is half of
.
Using the half-angle formula for tangent,
Solving the equation above, we get that or
. Since this value must be positive, we pick
. Then,
(since
is a right triangle with line
the diameter of the circumcircle) and
. Solving we get
,
, giving us a diagonal of length
and area
.
~Danielzh
Solution 3 (Analytic Geometry)
Denote by the half length of each side of the square.
We put the square to the coordinate plane, with
,
,
,
.
The radius of the circumcircle of is
.
Denote by
the argument of point
on the circle.
Thus, the coordinates of
are
.
Thus, the equations and
can be written as
These equations can be reformulated as
These equations can be reformulated as
Taking , by solving the equation, we get
Plugging (3) into (1), we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 (Law of Cosines)
WLOG, let be on minor arc
. Let
and
be the radius and center of the circumcircle respectively, and let
.
By the Pythagorean Theorem, the area of the square is . We can use the Law of Cosines on isosceles triangles
to get
Taking the products of the first two and last two equations, respectively, and
Adding these equations,
so
~OrangeQuail9
Solution 5 (Similar Triangles)
Someone please help render this
\begin{center}
\begin{tikzpicture} \draw (0,0) circle (4cm); \draw (2.8284, -2.8284) -- (2.8284, 2.8284) -- (-2.8284, 2.8284) -- (-2.8284, -2.8284) -- cycle; \draw (0, 0) node[anchor=north] {}; \draw (-2.8284, -2.8284) node[anchor=north east] {
}; \draw (2.8284, -2.8284) node[anchor=north west] {
}; \draw (2.8284, 2.8284) node[anchor=south west] {
}; \draw (-2.8284, 2.8284) node[anchor=south east] {
}; \draw (-0.531, 3.965) node[anchor=south] {
}; \draw (-2.8284, -2.8284) -- (2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; \draw (2.8284, -2.8284) -- (-2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; \draw[dashed] (-0.531, 3.965) -- (1.717, 1.717); \draw[dashed] (-0.531, 3.965) -- (-2.248, 2.248); \draw (-2.248, 2.248) node[anchor=north east] {
}; \draw (1.717, 1.717) node[anchor=north west] {
}; \end{tikzpicture}
\end{center}
Let the center of the circle be , and the radius of the circle be
. Since
is a rhombus with diagonals
and
, its area is
. Since
and
are diameters of the circle,
and
are right triangles. Let
and
be the foot of the altitudes to
and
, respectively. We have
so
. Similarly,
so
. Since
But
is a rectangle, so
, and our similarity becomes
Cross multiplying and rearranging gives us
, which rearranges to
. Therefore
.
~Cantalon
Solution 6 (Subtended Chords)
First draw a diagram.
Let's say that the radius is
. Then the area of the
is
Using the formula for the length of a chord subtended by an angle, we get
Multiplying and simplifying these 2 equations gives
Similarly
and
. Again, multiplying gives
Dividing
by
gives
, so
.
Pluging this back into one of the equations, gives
If we imagine a 28-45-53 right triangle, we see that if 28 is opposite and 45 is adjacent,
. Now we see that
~Voldemort101
Solution 7 (Areas and Pythagorean Theorem)
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.