Difference between revisions of "2024 AIME II Problems/Problem 12"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Question== | ||
+ | <asy> | ||
+ | pair O=(0,0); | ||
+ | pair X=(1,0); | ||
+ | pair Y=(0,1); | ||
+ | pair A=(0.5,0); pair B=(0,sin(pi/3)); | ||
+ | dot(O); | ||
+ | dot(X); | ||
+ | dot(Y); dot(A); dot(B); | ||
+ | draw(X--O--Y); | ||
+ | draw(A--B); | ||
+ | label("$B$", B, W); | ||
+ | pair P=(0.5, sin(pi/3)); | ||
+ | dot(P); | ||
+ | draw(A--P--B); | ||
+ | label("$A$", A, S); | ||
+ | label("$O$", O, SW); | ||
+ | pair C=(1/8,3*sqrt(3)/8); | ||
+ | dot(C); | ||
+ | label("$C$", C, SW); | ||
+ | draw(C--P); label("$P$", P, NE); | ||
+ | </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 so 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.) |
Revision as of 19:02, 9 February 2024
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)
Question
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 so 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.)