Difference between revisions of "2024 AIME II Problems/Problem 12"
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− | Let <math>O(0,0),A(\tfrac{1}{2},0),</math> and <math>B(0,\tfrac{\sqrt{3}}{2})</math> be points in the coordinate plane. Let <math>\mathcal{F}</math> be the family of segments <math>\overline{PQ}</math> of unit length lying in the first quadrant with <math>P</math> on the <math>x</math>-axis and <math>Q</math> on the <math>y</math>-axis. There is a unique point <math>C</math> on <math>\overline{AB},</math> distinct from <math>A</math> and <math>B,</math> that does not belong to any segment from <math>\mathcal{F}</math> other than <math>\overline{AB}</math>. Then <math>OC^2=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | + | ==Problem== |
+ | Let <math>O=(0,0),A=(\tfrac{1}{2},0),</math> and <math>B=(0,\tfrac{\sqrt{3}}{2})</math> be points in the coordinate plane. Let <math>\mathcal{F}</math> be the family of segments <math>\overline{PQ}</math> of unit length lying in the first quadrant with <math>P</math> on the <math>x</math>-axis and <math>Q</math> on the <math>y</math>-axis. There is a unique point <math>C</math> on <math>\overline{AB},</math> distinct from <math>A</math> and <math>B,</math> that does not belong to any segment from <math>\mathcal{F}</math> other than <math>\overline{AB}</math>. Then <math>OC^2=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
==Solution 1== | ==Solution 1== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == | + | ==Query== |
<asy> | <asy> | ||
pair O=(0,0); | pair O=(0,0); | ||
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</asy> | </asy> | ||
Let <math>C</math> be a fixed point in the first quadrant. Let <math>A</math> be a point on the positive <math>x</math>-axis and <math>B</math> be a point on the positive <math>y</math>-axis such that <math>AB</math> passes through <math>C</math> and the length of <math>AB</math> is minimal. Let <math>P</math> be the point such that <math>OAPB</math> is a rectangle. Prove that <math>PC \perp AB</math>. (One can solve this through algebra/calculus bash, but I'm trying to find a solution that mainly uses geometry. If you know such a solution, write it here on this wiki page.) ~Furaken | Let <math>C</math> be a fixed point in the first quadrant. Let <math>A</math> be a point on the positive <math>x</math>-axis and <math>B</math> be a point on the positive <math>y</math>-axis such that <math>AB</math> passes through <math>C</math> and the length of <math>AB</math> is minimal. Let <math>P</math> be the point such that <math>OAPB</math> is a rectangle. Prove that <math>PC \perp AB</math>. (One can solve this through algebra/calculus bash, but I'm trying to find a solution that mainly uses geometry. If you know such a solution, write it here on this wiki page.) ~Furaken | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2024|num-b=11|num-a=13|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:49, 9 February 2024
Problem
Let and
be points in the coordinate plane. Let
be the family of segments
of unit length lying in the first quadrant with
on the
-axis and
on the
-axis. There is a unique point
on
distinct from
and
that does not belong to any segment from
other than
. Then
, where
and
are relatively prime positive integers. Find
.
Solution 1
By Furaken
Let .
this is sus, furaken randomly guessed C and proceeded to prove it works Draw a line through intersecting the
-axis at
and the
-axis at
. We shall show that
, and that equality only holds when
and
.
Let . Draw
perpendicular to the
-axis and
perpendicular to the
-axis as shown in the diagram. Then
By some inequality (i forgor its name),
We know that
. Thus
. Equality holds if and only if
which occurs when
. Guess what,
happens to be
, thus
and
. Thus,
is the only segment in
that passes through
. Finally, we calculate
, and the answer is
.
~Furaken
Solution 2
When , the limit of
~Bluesoul
Solution 3
The equation of line is
\[
y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2)
\]
The position of line can be characterized by
, denoted as
.
Thus, the equation of line
is
\[
y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2)
\]
Solving (1) and (2), the -coordinate of the intersecting point of lines
and
satisfies the following equation:
\[
\frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta}
+ \frac{x}{\cos \theta}
= 1 . \hspace{1cm} (1)
\]
We denote the L.H.S. as
.
We observe that for all
.
Therefore, the point
that this problem asks us to find can be equivalently stated in the following way:
We interpret Equation (1) as a parameterized equation that is a tuning parameter and
is a variable that shall be solved and expressed in terms of
.
In Equation (1), there exists a unique
, denoted as
(
-coordinate of point
), such that the only solution is
. For all other
, there are more than one solutions with one solution
and at least another solution.
Given that function is differentiable, the above condition is equivalent to the first-order-condition
Calculating derivatives in this equation, we get
By solving this equation, we get
Plugging this into Equation (1), we get the -coordinate of point
:
Therefore, \begin{align*} OC^2 & = x_C^2 + y_C^2 \\ & = \frac{7}{16} . \end{align*}
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Query
Let
be a fixed point in the first quadrant. Let
be a point on the positive
-axis and
be a point on the positive
-axis such that
passes through
and the length of
is minimal. Let
be the point such that
is a rectangle. Prove that
. (One can solve this through algebra/calculus bash, but I'm trying to find a solution that mainly uses geometry. If you know such a solution, write it here on this wiki page.) ~Furaken
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.