Difference between revisions of "2023 AIME I Problems/Problem 5"
MRENTHUSIASM (talk | contribs) (→Solution 3 (Analytic geometry)) |
Voldemort101 (talk | contribs) (→Solution 6 (Similar Triangles)) |
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~Cantalon | ~Cantalon | ||
+ | ==Solution 7 (Subtended Chords)== | ||
+ | First draw a diagram. | ||
+ | <asy> | ||
+ | pair A, B, C, D, O, P; | ||
+ | A = (0,sqrt(106)); | ||
+ | B = (0,0); | ||
+ | C = (sqrt(106),0); | ||
+ | D = (sqrt(106),sqrt(106)); | ||
+ | O = (sqrt(106)/2, sqrt(106)/2); | ||
+ | P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(circle(O, sqrt(212)/2)); | ||
+ | label("$A$", A, NW); | ||
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE); | ||
+ | label("$D$", D, NE); | ||
+ | label("$P$", P, NW); | ||
+ | label("$O$", O, SW); | ||
+ | label("$\theta$", O, dir(120)*5); | ||
+ | dot(P); | ||
+ | dot(O); | ||
+ | draw(P--A--C--cycle, red); | ||
+ | draw(P--B--D--cycle, blue); | ||
+ | draw(P--O); | ||
+ | draw(anglemark(P,O,A,30)); | ||
+ | </asy> | ||
+ | Let's say that the radius is <math>r</math>. Then the area of the <math>ABCD</math> is <math>(\sqrt2r)^2 = 4r^2</math> | ||
+ | Using the formula for the length of a chord subtended by an angle, we get | ||
+ | <cmath>PA = 2r\sin\left(\dfrac{\theta}2\right)</cmath> | ||
+ | <cmath>PC = 2r\sin\left(\dfrac{180-\theta}2\right) = 2r\sin\left(90 - \dfrac{\theta}2\right) = 2r\cos\left(\dfrac{\theta}2\right)</cmath> | ||
+ | Multiplying and simplifying these 2 equations gives | ||
+ | <cmath>PA \cdot PC = 4r^2 \sin \left(\dfrac{\theta}2 \right) \cos \left(\dfrac{\theta}2 \right) = 2r^2 \sin\left(\theta \right) = 56</cmath> | ||
+ | Similarly <math>PB = 2r\sin\left(\dfrac{90 +\theta}2\right)</math> and <math>PD =2r\sin\left(\dfrac{90 -\theta}2\right)</math>. Again, multiplying gives | ||
+ | <cmath>PB \cdot PD = 4r^2 \sin\left(\dfrac{90 +\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right) = 4r^2 \sin\left(90 -\dfrac{90 -\theta}2\right) \sin\left(\dfrac{90 -\theta}2\right)</cmath> | ||
+ | <cmath> =4r^2 \sin\left(\dfrac{90 -\theta}2\right) \cos\left(\dfrac{90 -\theta}2\right) = 2r^2 \sin\left(90 - \theta \right) = 2r^2 \cos\left(\theta \right) = 90</cmath> | ||
+ | Dividing <math>2r^2 \sin \left(\theta \right)</math> by <math>2r^2 \cos \left( \theta \right)</math> gives <math>\tan \left(\theta \right) = \dfrac{28}{45}</math>, so <math>\theta = \tan^{-1} \left(\dfrac{28}{45} \right)</math>. | ||
+ | Pluging this back into one of the equations, gives | ||
+ | <cmath>2r^2 = \dfrac{90}{\cos\left(\tan^{-1}\left(\dfrac{28}{45}\right)\right)}</cmath> | ||
+ | If we imagine a 28-45-53 right triangle, we see that if 28 is opposite and 45 is adjacent, <math>\cos\left(\theta\right) = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{45}{53}</math>. Now we see that | ||
+ | <cmath>2r^2 = \dfrac{90}{\frac{45}{53}} = \boxed{106}</cmath> | ||
+ | ~ Voldemort101 | ||
==See also== | ==See also== |
Revision as of 17:29, 8 February 2023
Contents
Problem
Let be a point on the circle circumscribing square
that satisfies
and
Find the area of
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 (Double Angle)
Notice that and
are both
Solution 6 (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 7 (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
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.