Difference between revisions of "2024 AIME II Problems/Problem 12"
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− | Let's move a little bit from <math>A</math> to <math>A_1</math>, then <math>B</math> must move to <math>B_1</math> to keep <math>A_1B_1 = 1</math>. <math>AB</math> intersects with <math>A_1B_1</math> at <math>C</math>. Pick points <math>A_2</math> and <math>B_2</math> on <math>CA_1</math> and <math>CB</math> such that <math>CA_2 = CA</math>, <math>CB_2 = CB_1</math>, we have <math>A_1A_2 = BB_2</math>. Since <math>AA_1</math> is very small, <math>\angle CA_1A \approx 60^\circ</math>, <math>\angle CBB_1 \approx 30^\circ</math>, so <math>AA_2\approx \sqrt{3}A_1A_2</math>, <math>B_1B_2 \approx \frac{1}{\sqrt{3}}BB_2</math>, by similarity, <math>\frac{CA}{CB} \approx \frac{CA}{CB_2} = \frac{AA_2}{B_1B_2} = \frac{\sqrt{3} | + | Let's move a little bit from <math>A</math> to <math>A_1</math>, then <math>B</math> must move to <math>B_1</math> to keep <math>A_1B_1 = 1</math>. <math>AB</math> intersects with <math>A_1B_1</math> at <math>C</math>. Pick points <math>A_2</math> and <math>B_2</math> on <math>CA_1</math> and <math>CB</math> such that <math>CA_2 = CA</math>, <math>CB_2 = CB_1</math>, we have <math>A_1A_2 = BB_2</math>. Since <math>AA_1</math> is very small, <math>\angle CA_1A \approx 60^\circ</math>, <math>\angle CBB_1 \approx 30^\circ</math>, so <math>AA_2\approx \sqrt{3}A_1A_2</math>, <math>B_1B_2 \approx \frac{1}{\sqrt{3}}BB_2</math>, by similarity, <math>\frac{CA}{CB} \approx \frac{CA}{CB_2} = \frac{AA_2}{B_1B_2} = \frac{\sqrt{3}A_1A_3}{\frac{1}{\sqrt{3}}BB_2} = 3</math>. So the coordinates of <math>C</math> is <math>\left(\frac{1}{8}, \frac{3\sqrt{3}}{8}\right)</math>. |
so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>. | so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>. |
Revision as of 16:38, 13 February 2024
Contents
Problem
Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point \(C\) on \(\overline{AB}\), distinct from \(A\) and \(B\), that does not belong to any segment from \(\mathcal{F}\) other than \(\overline{AB}\). Then \(OC^2=\tfrac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).
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 forgot 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
Now, we want to find . By L'Hôpital's rule, we get
. This means that
, so we get
.
~Bluesoul
Solution 3
The equation of line is
The position of line can be characterized by
, denoted as
.
Thus, the equation of line
is
Solving (1) and (2), the -coordinate of the intersecting point of lines
and
satisfies the following equation:
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)
Solution 4 (coordinate bash)
Let be a segment in
with x-intercept
and y-intercept
. We can write
as
\begin{align*}
\frac{x}{a} + \frac{y}{b} &= 1 \\
y &= b(1 - \frac{x}{a}).
\end{align*}
Let the unique point in the first quadrant
lie on
and no other segment in
. We can find
by solving
and taking the limit as
. Since
has length
,
by the Pythagorean theorem. Solving this for
, we get
\begin{align*}
a^2 + b^2 &= 1 \\
b^2 &= 1 - a^2 \\
\frac{db^2}{da} &= \frac{d(1 - a^2)}{da} \\
2a\frac{db}{da} &= -2a \\
db &= -\frac{a}{b}da.
\end{align*}
After we substitute
, the equation for
becomes
In ,
and
. To find the x-coordinate of
, we substitute these into the equation for
and get
\begin{align*}
\frac{\sqrt{3}}{2}(1 - \frac{x}{\frac{1}{2}}) &= (\frac{\sqrt{3}}{2} - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} da)(1 - \frac{x}{\frac{1}{2} + da}) \\
\frac{\sqrt{3}}{2}(1 - 2x) &= (\frac{\sqrt{3}}{2} - \frac{da}{\sqrt{3}})(1 - \frac{x}{\frac{1 + 2da}{2}}) \\
\frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 - 2da}{2\sqrt{3}}(1 - \frac{2x}{1 + 2da}) \\
\frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 - 2da}{2\sqrt{3}} \cdot \frac{1 + 2da - 2x}{1 + 2da} \\
\frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 + 6da - 6x - 2da - 4da^2 + 4xda}{2\sqrt{3} + 4\sqrt{3}da} \\
(\frac{\sqrt{3}}{2} - \sqrt{3}x)(2\sqrt{3} + 4\sqrt{3}da) &= 3 + 6da - 6x - 2da - 4da^2 + 4xda \\
3 + 6da - 6x - 12xda &= 3 + 4da - 6x - 4da^2 + 4xda \\
2da &= -4da^2 + 16xda \\
16xda &= 2da + 4da^2 \\
x &= \frac{da + 2da^2}{8da}.
\end{align*}
We take the limit as
to get
We substitute
into the equation for
to find the y-coordinate of
:
The problem asks for
so
.
Solution 5 (small perturb)
Let's move a little bit from to
, then
must move to
to keep
.
intersects with
at
. Pick points
and
on
and
such that
,
, we have
. Since
is very small,
,
, so
,
, by similarity,
. So the coordinates of
is
.
so , the answer is
.
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.