Difference between revisions of "2024 AIME II Problems/Problem 12"

Revision as of 05:07, 9 February 2024

Let $O(0,0),A(\tfrac{1}{2},0),$ and $B(0,\tfrac{\sqrt{3}}{2})$ 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 $[asy] pair O=(0,0); pair X=(1,0); pair Y=(0,1); pair A=(0.5,0); pair B=(0,sin(pi/3)); dot(O); dot(X); dot(Y); dot(A); dot(B); draw(X--O--Y); draw(A--B); label("$B'$", B, W); label("$A'$", A, S); label("$O$", O, SW); pair C=(1/8,3*sqrt(3)/8); dot(C); pair D=(1/8,0); dot(D); pair E=(0,3*sqrt(3)/8); dot(E); label("$C$", C, NE); label("$D$", D, S); label("$E$", E, W); draw(D--C--E); [/asy]$

Let $C = (\tfrac18,\tfrac{3\sqrt3}8)$ . Draw a line through $C$ intersecting the $x$ -axis at $A'$ and the $y$ -axis at $B'$ . We shall show that $A'B' \ge 1$ , and that equality only holds when $A'=A$ and $B'=B$ .

Let $\theta = \angle OA'C$ . Draw $CD$ perpendicular to the $x$ -axis and $CE$ perpendicular to the $y$ -axis as shown in the diagram. Then $8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}$ By some inequality (i forgor its name), $\left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot \left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot (\sin^2\theta + \cos^2\theta) \ge (3+1)^3 = 64$ We know that $\sin^2\theta + \cos^2\theta = 1$ . Thus $\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8$ . Equality holds if and only if $\frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \sin^2\theta : \cos^2\theta$ which occurs when $\theta=\tfrac\pi3$ . Guess what, $\angle OAB$ happens to be $\tfrac\pi3$ , thus $A'=A$ and $B'=B$ . Thus, $AB$ is the only segment in $\mathcal{F}$ that passes through $C$ . Finally, we calculate $OC^2 = \tfrac1{64} + \tfrac{27}{64} = \tfrac7{16}$ , and the answer is $\boxed{023}$ .

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Difference between revisions of "2024 AIME II Problems/Problem 12"

Revision as of 05:07, 9 February 2024

Solution 1

@@ Line 1: / Line 1: @@
-:D :D :D
+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==
+By Furaken
+<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);
+label("$A'$", A, S);
+label("$O$", O, SW);
+pair C=(1/8,3*sqrt(3)/8);
+dot(C);
+pair D=(1/8,0); dot(D);
+pair E=(0,3*sqrt(3)/8); dot(E);
+label("$C$", C, NE); label("$D$", D, S); label("$E$", E, W);
+draw(D--C--E);
+</asy>
+Let <math>C = (\tfrac18,\tfrac{3\sqrt3}8)</math>. Draw a line through <math>C</math> intersecting the <math>x</math>-axis at <math>A'</math> and the <math>y</math>-axis at <math>B'</math>. We shall show that <math>A'B' \ge 1</math>, and that equality only holds when <math>A'=A</math> and <math>B'=B</math>.
+Let <math>\theta = \angle OA'C</math>. Draw <math>CD</math> perpendicular to the <math>x</math>-axis and <math>CE</math> perpendicular to the <math>y</math>-axis as shown in the diagram. Then
+<cmath>8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}</cmath>
+By some inequality (i forgor its name),
+<cmath>\left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot \left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot (\sin^2\theta + \cos^2\theta) \ge (3+1)^3 = 64</cmath>
+We know that <math>\sin^2\theta + \cos^2\theta = 1</math>. Thus <math>\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8</math>. Equality holds if and only if
+<cmath>\frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \sin^2\theta : \cos^2\theta</cmath>
+which occurs when <math>\theta=\tfrac\pi3</math>. Guess what, <math>\angle OAB</math> ''happens'' to be <math>\tfrac\pi3</math>, thus <math>A'=A</math> and <math>B'=B</math>. Thus, <math>AB</math> is the only segment in <math>\mathcal{F}</math> that passes through <math>C</math>. Finally, we calculate <math>OC^2 = \tfrac1{64} + \tfrac{27}{64} = \tfrac7{16}</math>, and the answer is <math>\boxed{023}</math>.