Difference between revisions of "2024 AIME II Problems/Problem 12"
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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 \(p+q\). | 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 \(p+q\). | ||
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− | Let <math>C = | + | ==Solution 1 (completely no calculus required)== |
+ | |||
+ | Begin by finding the equation of the line <math>\overline{AB}</math>: <math>y= -\sqrt{3}x+\frac{\sqrt{3}}{2}</math> Now, consider the general equation of all lines that belong to <math>\mathcal{F}</math>. Let <math>P</math> be located at <math>(a,0)</math> and <math>Q</math> be located at <math>(0,b)</math>. With these assumptions, we may arrive at the equation <math>ay +bx =ab</math>. However, a critical condition that must be satisfied by our parameters is that <math>a^2+b^2=1</math>, since the length of <math>\overline{PQ}=1</math>. | ||
+ | |||
+ | Here's the golden trick that resolves the problem: we wish to find some point <math>C</math> along <math>\overline{AB}</math> such that <math>\overline{PQ}</math> passes through <math>C</math> if <math>a=\frac{1}{2}</math>. It's not hard to convince oneself of this, since the property <math>a^2+b^2=1</math> implies that if <math>a=\frac{1}{2}</math>, then <math>\overline{PQ}=\overline{AB}</math>. | ||
+ | |||
+ | We should now try to relate the point <math>C</math> to some value of <math>a</math>. This is accomplished by finding the intersection of two lines: <cmath> | ||
+ | |||
+ | \[ | ||
+ | a(-\sqrt{3}x +\frac{\sqrt{3}}{2}) + x\sqrt{1-a^2} = a\sqrt{1-a^2} | ||
+ | \] | ||
+ | |||
+ | </cmath> | ||
+ | |||
+ | Where we have also used the fact that <math>b=\sqrt{1-a^2}</math>, which follows nicely from <math>a^2+b^2 =1</math>. <cmath> | ||
+ | |||
+ | \[ | ||
+ | a(-\sqrt{3}x +\frac{\sqrt{3}}{2}) = (a-x)\sqrt{1-a^2} | ||
+ | \] | ||
+ | |||
+ | </cmath> | ||
+ | |||
+ | Square both sides and go through some algebraic manipulations to arrive at | ||
+ | <cmath> | ||
+ | |||
+ | \[ | ||
+ | -a^4 +2xa^3+(-4x^2+3x+\frac{1}{4})a^2-2xa+x^2=0 | ||
+ | \] | ||
+ | |||
+ | </cmath> | ||
+ | |||
+ | Note how <math>a=\frac{1}{2}</math> is a solution to this polynomial, and it is logically so. If we found the set of intersections consisting of line segment <math>\overline{AB}</math> with an identical copy of itself, every single point on the line (all <math>x</math> values) should satisfy the equation. Thus, we can perform polynomial division to eliminate the extraneous solution <math>a=\frac{1}{2}</math>. <cmath> | ||
+ | |||
+ | \[ | ||
+ | -a^3 + (2x-\frac{1}{2})a^2+(-4x^2+4x)a-2x^2=0 | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Remember our original goal. It was to find an <math>x</math> value such that <math>a=\frac{1}{2}</math> is the only valid solution. Therefore, we can actually plug in <math>a=\frac{1}{2}</math> back into the equation to look for values of <math>x</math> such that the relation is satisfied, then eliminate undesirable answers. | ||
+ | <cmath> | ||
+ | |||
+ | \[ | ||
+ | 16x^2-10x+1=0 | ||
+ | \] | ||
+ | </cmath> | ||
+ | This is easily factored, allowing us to determine that <math>x=\frac{1}{8},\frac{1}{2}</math>. The latter root is not our answer, since on line <math>\overline{AB}</math>, <math>y(\frac{1}{2})=0</math>, the horizontal line segment running from <math>(0,0)</math> to <math>(1,0)</math> covers that point. From this, we see that <math>x=\frac{1}{8}</math> is the only possible candidate. | ||
+ | |||
+ | Going back to line <math>\overline{AB}, y= -\sqrt{3}x+\frac{\sqrt{3}}{2}</math>, plugging in <math>x=\frac{1}{8} </math> yields <math>y=\frac{3\sqrt{3}}{8}</math>. The distance from the origin is then given by <math>\sqrt{\frac{1}{8^2}+(\frac{3\sqrt{3}}{8})^2} =\sqrt{\frac{7}{16}}</math>. That number squared is <math>\frac{7}{16}</math>, so the answer is <math>\boxed{023}</math>. | ||
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+ | |||
− | + | ~Installhelp_hex | |
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− | ~ | ||
==Solution 2== | ==Solution 2== | ||
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<math>y=-(\tan \theta) x+\sin \theta=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math> | <math>y=-(\tan \theta) x+\sin \theta=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math> | ||
− | Now, we want to find <math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math>. By L'Hôpital's rule, we get <math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{ | + | Now, we want to find <math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math>. By L'Hôpital's rule, we get <math>x=\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{\theta}=\frac{1}{8}</math>. This means that <math>y=\frac{3\sqrt{3}}{8}\implies OC^2=\frac{7}{16}</math>, so we get <math>\boxed{023}</math>. |
~Bluesoul | ~Bluesoul | ||
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</asy> | </asy> | ||
− | 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_2}{\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>. | ||
+ | |||
+ | ==Solution 6(trig identities and questionable rigidity)== | ||
+ | |||
+ | Let's try to find the general form of a line that is in <math>\mathcal{F}</math> based on what angle it makes with the x-axis, <math>OP = \cos{\theta}</math>, and <math>OQ = \sin{\theta}</math> so its slope is <math>-\tan{\theta}</math> and due to us knowing its y-intercept we know that our line has form <math>y = -\tan{\theta}x + \sin{\theta}.</math> | ||
+ | |||
+ | Now we can analyze the system of equations made by <math>y = -\tan{\theta}x + \sin{\theta}</math> and <math>y = -\tan{60^\circ}x + \sin{60^\circ}</math>, this gives us that <math>x = \dfrac{\sin{\theta}- \sin{60^\circ}}{\tan{\theta} - \sin{60^\circ}}.</math> | ||
+ | |||
+ | We can proceed to simplify our expression further: | ||
+ | <cmath> x = \dfrac{\sin{\theta} - \sin{60^\circ}}{\frac{\sin{\theta}}{\cos{\theta}} - \frac{\sin{60^\circ}}{\cos{60^\circ}}}</cmath> <cmath>= \dfrac{\sin{\theta} - \sin{60^\circ}}{\frac{\sin{(\theta - 60^\circ)}}{\cos{\theta}\cos{60^\circ}}}</cmath> <cmath>= \dfrac{2\sin{\dfrac{\theta - 60^\circ}{2}}\cos{\dfrac{\theta + 60^\circ}{2}}}{\dfrac{2\sin{\dfrac{\theta - 60^\circ}{2}}\cos{\dfrac{\theta - 60^\circ}{2}}}{\cos{\theta}\cos{60^\circ}}} </cmath> <cmath> = \dfrac{\sin{\dfrac{\theta - 60^\circ}{2}}}{\sin{\dfrac{\theta - 60^\circ}{2}}} \cdot \dfrac{\cos{\theta}\cos{60^\circ}\cos{\dfrac{\theta + 60^\circ}{2}}}{\cos{\dfrac{\theta - 60^\circ}{2}}}.</cmath> | ||
+ | |||
+ | Seeing that there are only valid solutions when <math>\theta</math> is acute(all that is allowed anyways) and when <math>\theta \neq 60^\circ</math> since one of the expressions in our simplified solution will equal <math>0/0</math>. Since there is only one intersection point for every <math>\theta</math> and vice versa in the appropriate domain and range(we can easily prove this by contradiction), we know that the missing element of the range(the points) must correspond with the excluded value. The x-coordinate of which which can be evaluated by taking the limit of our expression as <math>\theta</math> goes to <math>60^\circ</math> which is <math>\frac{1}{8}</math> regardless of the direction we approach <math>60^\circ</math> from. The corresponding <math>y</math> is <math>\dfrac{3\sqrt{3}}{8}</math> and using the distance formula gives us <math>\boxed{023}</math> as our answer. | ||
+ | |||
+ | While this solution may seem long all of these steps come naturally. | ||
+ | |||
+ | ~SailS | ||
+ | ==Video Solution | ||
+ | |||
+ | https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL | ||
+ | |||
+ | (no calculus) | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
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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 | ||
+ | |||
+ | I think there is such a geometry way: | ||
+ | Let <math>DE</math> pass through <math>C</math> while point <math>D</math> is on the outside of line segment <math>OA</math> and point <math>E</math> is in between <math>O</math> and <math>B</math>. We aim to show <math>DE</math> is longer than <math>AB</math>. Now since <math>PC</math> is the altitude of triangle <math>PAB</math> yet just a cevian on the base <math>DE</math> of triangle <math>PDE</math> (thus making the height shorter than <math>PC</math>), it suffices to show the area of triangle <math>PDE</math> is bigger than that of triangle <math>PAB</math>. To do this, we compare these two triangles (let <math>DE</math> intersect <math>PA</math> at point <math>F</math>), and we just want to show <math>PF*AD > AF*AO</math>. This is trivial by similarity ratios. ~gougutheorem | ||
+ | |||
+ | Thanks! Now we know that it's possible to solve the AIME problem with only geometry. ~Furaken | ||
==See also== | ==See also== |
Latest revision as of 09:06, 25 June 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 (completely no calculus required)
Begin by finding the equation of the line :
Now, consider the general equation of all lines that belong to
. Let
be located at
and
be located at
. With these assumptions, we may arrive at the equation
. However, a critical condition that must be satisfied by our parameters is that
, since the length of
.
Here's the golden trick that resolves the problem: we wish to find some point along
such that
passes through
if
. It's not hard to convince oneself of this, since the property
implies that if
, then
.
We should now try to relate the point to some value of
. This is accomplished by finding the intersection of two lines:
Where we have also used the fact that , which follows nicely from
.
Square both sides and go through some algebraic manipulations to arrive at
Note how is a solution to this polynomial, and it is logically so. If we found the set of intersections consisting of line segment
with an identical copy of itself, every single point on the line (all
values) should satisfy the equation. Thus, we can perform polynomial division to eliminate the extraneous solution
.
Remember our original goal. It was to find an value such that
is the only valid solution. Therefore, we can actually plug in
back into the equation to look for values of
such that the relation is satisfied, then eliminate undesirable answers.
This is easily factored, allowing us to determine that
. The latter root is not our answer, since on line
,
, the horizontal line segment running from
to
covers that point. From this, we see that
is the only possible candidate.
Going back to line , plugging in
yields
. The distance from the origin is then given by
. That number squared is
, so the answer is
.
~Installhelp_hex
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
.
Solution 6(trig identities and questionable rigidity)
Let's try to find the general form of a line that is in based on what angle it makes with the x-axis,
, and
so its slope is
and due to us knowing its y-intercept we know that our line has form
Now we can analyze the system of equations made by and
, this gives us that
We can proceed to simplify our expression further:
Seeing that there are only valid solutions when is acute(all that is allowed anyways) and when
since one of the expressions in our simplified solution will equal
. Since there is only one intersection point for every
and vice versa in the appropriate domain and range(we can easily prove this by contradiction), we know that the missing element of the range(the points) must correspond with the excluded value. The x-coordinate of which which can be evaluated by taking the limit of our expression as
goes to
which is
regardless of the direction we approach
from. The corresponding
is
and using the distance formula gives us
as our answer.
While this solution may seem long all of these steps come naturally.
~SailS ==Video Solution
https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL
(no calculus)
~MathProblemSolvingSkills.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
I think there is such a geometry way:
Let pass through
while point
is on the outside of line segment
and point
is in between
and
. We aim to show
is longer than
. Now since
is the altitude of triangle
yet just a cevian on the base
of triangle
(thus making the height shorter than
), it suffices to show the area of triangle
is bigger than that of triangle
. To do this, we compare these two triangles (let
intersect
at point
), and we just want to show
. This is trivial by similarity ratios. ~gougutheorem
Thanks! Now we know that it's possible to solve the AIME problem with only geometry. ~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.