Difference between revisions of "2024 AIME II Problems/Problem 12"
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− | + | ==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 and only 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>. | ||
+ | |||
+ | |||
− | + | ~Installhelp_hex | |
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− | ~ | ||
==Solution 2== | ==Solution 2== | ||
− | <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>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 | ||
Line 46: | Line 66: | ||
==Solution 3== | ==Solution 3== | ||
− | The equation of line <math>AB</math> is | + | The equation of line <math>AB</math> is <cmath> |
\[ | \[ | ||
− | y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} ( | + | y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (1) |
\] | \] | ||
+ | |||
+ | </cmath> | ||
The position of line <math>PQ</math> can be characterized by <math>\angle QPO</math>, denoted as <math>\theta</math>. | The position of line <math>PQ</math> can be characterized by <math>\angle QPO</math>, denoted as <math>\theta</math>. | ||
Thus, the equation of line <math>PQ</math> is | Thus, the equation of line <math>PQ</math> is | ||
+ | |||
+ | <cmath> | ||
\[ | \[ | ||
y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) | y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) | ||
\] | \] | ||
+ | </cmath> | ||
Solving (1) and (2), the <math>x</math>-coordinate of the intersecting point of lines <math>AB</math> and <math>PQ</math> satisfies the following equation: | Solving (1) and (2), the <math>x</math>-coordinate of the intersecting point of lines <math>AB</math> and <math>PQ</math> satisfies the following equation: | ||
+ | |||
+ | <cmath> | ||
\[ | \[ | ||
\frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} | \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} | ||
Line 63: | Line 90: | ||
= 1 . \hspace{1cm} (1) | = 1 . \hspace{1cm} (1) | ||
\] | \] | ||
− | We denote the L.H.S. as <math>f \left( \theta ; x \right)</math>. | + | </cmath> |
+ | |||
+ | We denote the L.H.S. as <math>f \left( \theta; x \right)</math>. | ||
We observe that <math>f \left( 60^\circ ; x \right) = 1</math> for all <math>x</math>. | We observe that <math>f \left( 60^\circ ; x \right) = 1</math> for all <math>x</math>. | ||
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- \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} | - \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} | ||
+ x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} | + x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} | ||
− | = 0 | + | = 0. |
\] | \] | ||
</cmath> | </cmath> | ||
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& = \frac{7}{16} . | & = \frac{7}{16} . | ||
\end{align*} | \end{align*} | ||
− | |||
Therefore, the answer is <math>7 + 16 = \boxed{\textbf{(23) }}</math>. | Therefore, the answer is <math>7 + 16 = \boxed{\textbf{(23) }}</math>. | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | == Solution 4 (coordinate bash) == | ||
+ | |||
+ | Let <math>s</math> be a segment in <math>\mathcal{F}</math> with x-intercept <math>a</math> and y-intercept <math>b</math>. We can write <math>s</math> 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 <math>(x, y)</math> lie on <math>s</math> and no other segment in <math>\mathcal{F}</math>. We can find <math>x</math> by solving | ||
+ | <cmath> | ||
+ | b(1 - \frac{x}{a}) = (b + db)(1 - \frac{x}{a + da}) | ||
+ | </cmath> | ||
+ | and taking the limit as <math>da, db \to 0</math>. Since <math>s</math> has length <math>1</math>, <math>a^2 + b^2 = 1^2</math> by the Pythagorean theorem. Solving this for <math>db</math>, 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 <math>db = -\frac{a}{b}da</math>, the equation for <math>x</math> becomes | ||
+ | <cmath> | ||
+ | b(1 - \frac{x}{a}) = (b -\frac{a}{b} da)(1 - \frac{x}{a + da}). | ||
+ | </cmath> | ||
+ | |||
+ | In <math>\overline{AB}</math>, <math>a = \frac{1}{2}</math> and <math>b = \frac{\sqrt{3}}{2}</math>. To find the x-coordinate of <math>C</math>, we substitute these into the equation for <math>x</math> 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 <math>da \to 0</math> to get | ||
+ | <cmath> | ||
+ | x = \lim_{da \to 0} \frac{da + 2da^2}{8da} = \lim_{da \to 0} \frac{1 + 2da}{8} = \frac{1}{8}. | ||
+ | </cmath> | ||
+ | We substitute <math>x = \frac{1}{8}</math> into the equation for <math>\overline{AB}</math> to find the y-coordinate of <math>C</math>: | ||
+ | <cmath> | ||
+ | y = b(1 - \frac{x}{a}) = \frac{\sqrt{3}}{2}(1 - \frac{\frac{1}{8}}{\frac{1}{2}}) = \frac{3\sqrt{3}}{8}. | ||
+ | </cmath> | ||
+ | The problem asks for | ||
+ | <cmath> | ||
+ | OC^2 = x^2 + y^2 = (\frac{1}{8})^2 + (\frac{3\sqrt{3}}{8})^2 = \frac{7}{16} = \frac{p}{q}, | ||
+ | </cmath> | ||
+ | so <math>p + q = 7 + 16 = \boxed{023}</math>. | ||
+ | |||
+ | == Solution 5 (small perturb) == | ||
+ | |||
+ | <asy> | ||
+ | pair O=(0,0); | ||
+ | pair X=(1,0); | ||
+ | pair Y=(0,1); | ||
+ | pair A=(0.5,0); pair B=(0,sin(pi/3)); | ||
+ | pair A1=(0.6,0); pair B1=(0,0.8); | ||
+ | pair A2=(0.575,0.04); pair B2=(0.03,0.816); | ||
+ | dot(O); | ||
+ | dot(X); | ||
+ | dot(Y); dot(A); dot(B); | ||
+ | dot(A1); dot(B1); | ||
+ | dot(A2); dot(B2); | ||
+ | draw(X--O--Y); | ||
+ | draw(A--B); | ||
+ | draw(A1--B1); | ||
+ | draw(A--A2); | ||
+ | draw(B1--B2); | ||
+ | label("$B$", B, W); | ||
+ | label("$A$", A, S); | ||
+ | label("$B_1$", B1, SW); | ||
+ | label("$A_1$", A1, S); | ||
+ | label("$B_2$", B2, E); | ||
+ | label("$A_2$", A2, NE); | ||
+ | label("$O$", O, SW); | ||
+ | pair C=(0.18,0.56); | ||
+ | label("$C$", C, E); | ||
+ | dot(C); | ||
+ | </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}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>. | ||
+ | |||
+ | ==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 | ||
+ | ==Solution 7== | ||
+ | [[File:2024 AIME II 12 a.png|330px|right]] | ||
+ | Denote <math>X(1,0), Y(0,1), P \in OX, Q \in OY, |PQ| = 1,</math> | ||
+ | <cmath>k = const \in (0,1), \alpha = \angle OPQ.</cmath> | ||
+ | Then <math>P = (\cos \alpha, 0), Q = (0, \sin \alpha).</math> | ||
+ | |||
+ | Let <math>C \in PQ</math> be the point with property <math>k = \frac {PC}{QC} \implies </math> | ||
+ | <math>C = \left ( \frac{\cos \alpha}{k+1} , \frac{k \sin \alpha}{k+1} \right ).</math> | ||
+ | |||
+ | So locus of points <math>C</math> is the ellipse with semiaxes <math>\frac{1}{k+1}</math> and <math>\frac{k}{k+1}.</math> | ||
+ | |||
+ | Point <math>D = C</math> is a unique point on <math>AB</math> if the ellipse is tangent to the line <math>AB.</math> | ||
+ | |||
+ | In this case in point <math>C</math> we get <math>\frac{dy}{dx} = \frac{dy}{d \alpha} : \frac{dx}{d \alpha} = - k \cot \alpha.</math> | ||
+ | |||
+ | The tangent of the line <math>AB</math> is <math>- \tan \alpha \implies k = \tan^2 \alpha.</math> | ||
+ | |||
+ | For point <math>D</math> we get <math>k = \tan^2 \theta.</math> | ||
+ | |||
+ | <cmath>\tan \angle DOX = \tan \varphi = \frac {D_y}{D_x} = \frac{k \sin \theta}{k+1} : \frac{\cos \theta}{k+1} = k \cdot \tan \theta = \tan^3 \theta.</cmath> | ||
+ | |||
+ | For the line <math>AB \tan \theta = \frac {B_y}{A_x} = \sqrt{3} \implies k = 3 \implies</math> | ||
+ | <cmath>D = \left( \frac{\cos \theta}{k+1}, \frac{k \sin \theta}{k+1} \right) = | ||
+ | \left( \frac{1}{\sqrt{1+ \tan^2 \theta} \cdot (k+1)}, \frac{k \tan \theta}{\sqrt{1+ \tan^2 \theta} \cdot (k+1)} \right) =</cmath> | ||
+ | <math> \left(\frac{1}{\sqrt{1+ k} \cdot (k+1)}, \frac{k \cdot \sqrt{k}}{\sqrt{1+ k} \cdot (k+1)} \right) = \left(\frac{1}{8}, \frac{3 \sqrt{3}}{8} \right),</math> so we get <math>\boxed{023}</math>. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 8 (intuitive elementary calculus solution)== | ||
+ | |||
+ | First, we find the equation of the line <math>\overline{AB}:</math> | ||
+ | <cmath>\[ | ||
+ | y = -\sqrt{3} x + \dfrac{\sqrt{3}}{2}. \hspace{1cm} (1) | ||
+ | \]</cmath> | ||
+ | Now, say we have some line in <math>\mathcal{F}</math> that spans from point <math>X</math> at <math>(0,a)</math>, and because this segment has unit length, ends at point <math>Y</math> located at <math>(\sqrt{1-a^2},0)</math>. Note that <math>\overline{XY}</math> has the equation | ||
+ | <cmath>\[ | ||
+ | y = -\dfrac{a}{\sqrt{1-a^2}}x + a. \hspace{1cm} (2) | ||
+ | \] </cmath> | ||
+ | We notice that as <math>a</math> varies, <math>\overline{XY}</math> will intersect <math>\overline{AB}</math> at all points on <math>\overline{AB}</math> except point <math>C</math>. We make the key observation that <math>C</math> is what the intersection point of <math>\overline{XY}</math> and <math>\overline{AB}</math> approaches as <math>a</math> approaches <math>\dfrac{\sqrt{3}}{2}</math>. | ||
+ | |||
+ | To find this point, let's find the intersection point of <math>\overline{XY}</math> and <math>\overline{AB}</math> in terms of <math>a</math>. | ||
+ | \begin{align*} | ||
+ | y &= -\sqrt{3} x + \dfrac{\sqrt{3}}{2} \ | ||
+ | y &= -\dfrac{a}{\sqrt{1-a^2}}x + a \ | ||
+ | -\sqrt{3} x + \dfrac{\sqrt{3}}{2} &= -\dfrac{a}{\sqrt{1-a^2}}x + a \ | ||
+ | (\dfrac{a}{\sqrt{1-a^2}} - \sqrt{3}) \cdot x &= a-\dfrac{\sqrt{3}}{2} \ | ||
+ | x &= \dfrac{(a-\dfrac{\sqrt{3}}{2})\cdot\sqrt{1-a^2}}{a - \sqrt{3} \cdot \sqrt{1-a^2}} \ | ||
+ | \end{align*} | ||
+ | |||
+ | Note that we don't need to find the <math>y</math>-coordinate because we know that this point is on line <math>\overline{AB}</math>, and once we find the <math>x</math>-coordinate, we can simply plug it into the equation of line <math>\overline{AB}</math> | ||
+ | |||
+ | Now, we want to find <cmath> | ||
+ | \[ | ||
+ | \lim_{a\to\frac{\sqrt{3}}{2}} \dfrac{(a-\dfrac{\sqrt{3}}{2})\cdot\sqrt{1-a^2}}{a - \sqrt{3} \cdot \sqrt{1-a^2}} | ||
+ | \] | ||
+ | </cmath> | ||
+ | We apply L'Hopital's Rule as this limit is indeterminate. | ||
+ | Taking the derivative of the numerator using product rule: | ||
+ | \begin{align*} | ||
+ | \dfrac{d}{da} \left[(a-\dfrac{\sqrt{3}}{2})\cdot\sqrt{1-a^2}\right] &= \sqrt{1-a^2} \cdot \dfrac{d}{da} \left(a-\dfrac{\sqrt{3}}{2}\right) + \dfrac{d}{da} \left(\sqrt{1-a^2}\right) \cdot (a-\dfrac{\sqrt{3}}{2}) | ||
+ | \end{align*} | ||
+ | Note that we can apply the chain rule to get <math>\dfrac{d}{da} \left(\sqrt{1-a^2}\right) = \dfrac{-a}{\sqrt{1-a^2}}</math> | ||
+ | \begin{align*} | ||
+ | &= \sqrt{1-a^2} + \dfrac{-a}{\sqrt{1-a^2}} * (a-\dfrac{\sqrt{3}}{2}) \ | ||
+ | &= \dfrac{-2a^2 + \dfrac{a\sqrt{3}}{2} + 1}{\sqrt{1-a^2}} | ||
+ | \end{align*} | ||
+ | Taking the derivative of the denominator: | ||
+ | \begin{align*} | ||
+ | \dfrac{d}{da} \left[a-\sqrt{3}\cdot\sqrt{1-a^2}\right] &= \dfrac{d}{da} (a) - \sqrt{3}\dfrac{d}{da}(\sqrt{1-a^2}) \ | ||
+ | &= 1 - \sqrt{3}\cdot\dfrac{-a}{\sqrt{1-a^2}} \ | ||
+ | &= \dfrac{\sqrt{1-a^2} + \sqrt{3}a}{\sqrt{1-a^2}} | ||
+ | \end{align*} | ||
+ | So, our final expression is <cmath> | ||
+ | \[ | ||
+ | \dfrac{\frac{-2a^2 + \frac{a\sqrt{3}}{2} + 1}{\sqrt{1-a^2}}}{\frac{\sqrt{1-a^2} + \sqrt{3}a}{\sqrt{1-a^2}}} | ||
+ | \] </cmath> | ||
+ | <cmath> | ||
+ | \[ | ||
+ | = \dfrac{-2a^2 + \frac{a\sqrt{3}}{2} + 1}{\sqrt{1-a^2} + \sqrt{3}a} | ||
+ | \]</cmath> | ||
+ | |||
+ | Now, all that remains is to substitute <math>a = \dfrac{\sqrt{3}}{2}</math> | ||
+ | \begin{align*} | ||
+ | &= \dfrac{-\dfrac{3}{2} + \dfrac{3}{4} + 1}{\dfrac{1}{2} + \dfrac{3}{2}} \ | ||
+ | &= \dfrac{1}{8} | ||
+ | \end{align*} | ||
+ | Now, we can plug in <math>x = \dfrac{1}{8}</math> into the equation of <math>\overline{AB}</math>: | ||
+ | \begin{align*} | ||
+ | y &= -\dfrac{\sqrt{3}}{8} + \dfrac{\sqrt{3}}{2} \ | ||
+ | y &= \dfrac{3\sqrt{3}}{8} | ||
+ | \end{align*} | ||
+ | So point C is located at <math>(\dfrac{1}{8},\dfrac{3\sqrt{3}}{8})</math> | ||
+ | The question asks for <math>OC^2</math>, so we simply apply the Pythagorean Theorem: | ||
+ | \begin{align*} | ||
+ | OC^2 &= (\dfrac{1}{8})^2 + (\dfrac{3\sqrt{3}}{2})^2 \ | ||
+ | OC^2 &= \dfrac{7}{16}\ | ||
+ | 7 + 16 &= \boxed{23} | ||
+ | \end{align*} | ||
+ | |||
+ | ~143466534811009我输了游戏56二伯 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL | ||
+ | |||
+ | (no calculus) | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==Video Solution== | ==Video Solution== | ||
Line 141: | Line 378: | ||
</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== |
Revision as of 17:56, 17 November 2024
Contents
[hide]- 1 Problem
- 2 Solution 1 (completely no calculus required)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (coordinate bash)
- 6 Solution 5 (small perturb)
- 7 Solution 6(trig identities and questionable rigidity)
- 8 Solution 7
- 9 Solution 8 (intuitive elementary calculus solution)
- 10 Video Solution
- 11 Video Solution
- 12 Query
- 13 See also
Problem
Let
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 and only 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,
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
In , and . To find the x-coordinate of , we substitute these into the equation for and get
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
Solution 7
Denote Then
Let be the point with property
So locus of points is the ellipse with semiaxes and
Point is a unique point on if the ellipse is tangent to the line
In this case in point we get
The tangent of the line is
For point we get
For the line so we get .
vladimir.shelomovskii@gmail.com, vvsss
Solution 8 (intuitive elementary calculus solution)
First, we find the equation of the line Now, say we have some line in that spans from point at , and because this segment has unit length, ends at point located at . Note that has the equation We notice that as varies, will intersect at all points on except point . We make the key observation that is what the intersection point of and approaches as approaches .
To find this point, let's find the intersection point of and in terms of .
Note that we don't need to find the -coordinate because we know that this point is on line , and once we find the -coordinate, we can simply plug it into the equation of line
Now, we want to find
We apply L'Hopital's Rule as this limit is indeterminate.
Taking the derivative of the numerator using product rule:
Now, all that remains is to substitute
~143466534811009我输了游戏56二伯
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.