Difference between revisions of "2024 AMC 12A Problems/Problem 24"

Revision as of 21:36, 20 March 2025

Problem

There exist exactly four different complex numbers $z$ that satisfy the equation shown below: $\left|z + \overline{z}\left(\tfrac{3}{5} + \tfrac{4}{5}i\right)\right| = \left|z + \overline{z}\left(\tfrac{4}{5} + \tfrac{3}{5}i\right)\right| = 1.$ What is the area of the convex quadrilateral whose vertices are those four complex numbers $z$ in the complex plane?

$\textbf{(A)}~\frac{4\sqrt{21}}{3}\qquad\textbf{(B)}~\frac{9\sqrt{2}}{2}\qquad\textbf{(C)}~5\sqrt{2}\qquad\textbf{(D)}~2\sqrt{14}\qquad\textbf{(E)}~\frac{7\sqrt{5}}{2}$

Solution

Let $ABCD$ be the rectangle with vertices $\pm\sqrt{\tfrac{3}{5}+\tfrac{4}{5}i}$ and $\pm\sqrt{\tfrac{4}{5}+\tfrac{3}{5}i}$ and let $O$ be its center. Let the reflections of $z$ over $\overline{AC}$ and $\overline{BD}$ be $z_1$ and $z_2$ . If the points $X$ and $Y$ are such that $OzXz_1$ and $OzYz_2$ are parallelograms, then the given condition is equivalent to $X$ and $Y$ lying on $(ABCD)$ . The condition that $X$ lies on this circle is equivalent to $z$ lying on the union of the perpendicular bisectors of $\overline{OA}$ and $\overline{OC}$ and the condition that $Y$ lies on the circle is equivalent to $z$ lying on the union of the perpendicular bisectors of $\overline{OB}$ and $\overline{OD}$ . Therefore, the possible $z$ are the circumcenters of $\triangle{}OAB,\triangle{}OBC,\triangle{}OCD,$ and $\triangle{}ODA$ and therefore they form a rhombus. Now, the area of this rhombus is twice the product of the circumradii of $\triangle{}OAB$ and $\triangle{}OBC$ , and if $a=\angle{}AOB$ then this is just $2\cdot\tfrac{1}{2\sin\left(90^\circ-\frac{a}{2}\right)}\cdot\tfrac{1}{2\sin\left(90^\circ-\frac{180^\circ-a}{2}\right)}=\tfrac{1}{\sin(a)}$ . Now, we can find that $\tfrac{\frac{3}{5}+\frac{4}{5}i}{\frac{4}{5}+\frac{3}{5}i}=\tfrac{24}{25}+\tfrac{7}{25}i$ so $\cos(2a)=\tfrac{24}{25}$ and $\sin(a)=\sqrt{\tfrac{1-\frac{24}{25}}{2}}=\tfrac{\sqrt{2}}{10}$ , giving that the answer is $\boxed{\textbf{(C)}\ 5\sqrt{2}}$ .

~Zhaom

@@ Line 1: / Line 1: @@
 ==Problem==
-A <math>\textit{disphenoid}</math> is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?
+There exist exactly four different complex numbers <math>z</math> that satisfy the equation shown below: <cmath>\left|z + \overline{z}\left(\tfrac{3}{5} + \tfrac{4}{5}i\right)\right| = \left|z + \overline{z}\left(\tfrac{4}{5} + \tfrac{3}{5}i\right)\right| = 1.</cmath> What is the area of the convex quadrilateral whose vertices are those four complex numbers <math>z</math> in the complex plane?
-<math>\textbf{(A) }\sqrt{3}\qquad\textbf{(B) }3\sqrt{15}\qquad\textbf{(C) }15\qquad\textbf{(D) }15\sqrt{7}\qquad\textbf{(E) }24\sqrt{6}</math>
+<math>\textbf{(A)}~\frac{4\sqrt{21}}{3}\qquad\textbf{(B)}~\frac{9\sqrt{2}}{2}\qquad\textbf{(C)}~5\sqrt{2}\qquad\textbf{(D)}~2\sqrt{14}\qquad\textbf{(E)}~\frac{7\sqrt{5}}{2}</math>
-==Solution 1 (Definition of disphenoid)==
+==Solution==
+Let <math>ABCD</math> be the rectangle with vertices <math>\pm\sqrt{\tfrac{3}{5}+\tfrac{4}{5}i}</math> and <math>\pm\sqrt{\tfrac{4}{5}+\tfrac{3}{5}i}</math> and let <math>O</math> be its center. Let the reflections of <math>z</math> over <math>\overline{AC}</math> and <math>\overline{BD}</math> be <math>z_1</math> and <math>z_2</math>. If the points <math>X</math> and <math>Y</math> are such that <math>OzXz_1</math> and <math>OzYz_2</math> are parallelograms, then the given condition is equivalent to <math>X</math> and <math>Y</math> lying on <math>(ABCD)</math>. The condition that <math>X</math> lies on this circle is equivalent to <math>z</math> lying on the union of the perpendicular bisectors of <math>\overline{OA}</math> and <math>\overline{OC}</math> and the condition that <math>Y</math> lies on the circle is equivalent to <math>z</math> lying on the union of the perpendicular bisectors of <math>\overline{OB}</math> and <math>\overline{OD}</math>. Therefore, the possible <math>z</math> are the circumcenters of <math>\triangle{}OAB,\triangle{}OBC,\triangle{}OCD,</math> and <math>\triangle{}ODA</math> and therefore they form a rhombus. Now, the area of this rhombus is twice the product of the circumradii of <math>\triangle{}OAB</math> and <math>\triangle{}OBC</math>, and if <math>a=\angle{}AOB</math> then this is just <math>2\cdot\tfrac{1}{2\sin\left(90^\circ-\frac{a}{2}\right)}\cdot\tfrac{1}{2\sin\left(90^\circ-\frac{180^\circ-a}{2}\right)}=\tfrac{1}{\sin(a)}</math>. Now, we can find that <math>\tfrac{\frac{3}{5}+\frac{4}{5}i}{\frac{4}{5}+\frac{3}{5}i}=\tfrac{24}{25}+\tfrac{7}{25}i</math> so <math>\cos(2a)=\tfrac{24}{25}</math> and <math>\sin(a)=\sqrt{\tfrac{1-\frac{24}{25}}{2}}=\tfrac{\sqrt{2}}{10}</math>, giving that the answer is <math>\boxed{\textbf{(C)}\ 5\sqrt{2}}</math>.
-Notice that any scalene <math>\textit{acute}</math> triangle can be the faces of a <math>\textit{disphenoid}</math>. (See proof in Solution 2.)
+~Zhaom
-As a result, we simply have to find the smallest area a scalene acute triangle with integer side lengths can take on. This occurs with a <math>4,5,6</math> triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles, degenerate, or non-acute). For this triangle, the semiperimeter is <math>\frac{15}{2}</math>, so by Heron’s Formula:
-\begin{align*}
-A&=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\\
-&=\sqrt{\frac{15^2\cdot7}{16}}\\
-&=\frac{15}{4}\sqrt{7}
-\end{align*}
-The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
-~eevee9406
-==Solution 2 (Disphenoid in Box)==
-Let the side lengths of one face of the disphenoid be <math>a, b, c</math>. By the definition of a disphenoid with scalene faces, opposite sides must be the same length. Then the disphenoid can be constructed in a rectangular box with dimensions <math>p, q, r</math> such that <math>a, b, c</math> are the <math>3</math> different diagonal lengths of the faces of the box and no two sides are parallel (someone feel free to insert a diagram). Then we have the system
-<cmath>p^2 + q^2 = a^2</cmath>
-<cmath>p^2 + r^2 = b^2</cmath>
-<cmath>q^2 + r^2 = c^2</cmath>
-for positive integers <math>a, b, c</math> and positive <math>p, q, r</math>.
-Solving for <math>p, q, r</math>, we have
-<cmath>p^2 = \frac{a^2 + b^2 - c^2}{2}</cmath>
-<cmath>q^2 = \frac{a^2 - b^2 + c^2}{2}</cmath>
-<cmath>r^2 = \frac{-a^2 + b^2 + c^2}{2}</cmath>
-(Notice that, by law of cosines (or by Pythagorean Inequality), these (parallelepiped box side lengths squared) <math>p^2, q^2, r^2</math> each have the same sign as the cosine of the angle at a vertex of a triangular face of the disphenoid. Since they are squares and so non-negative, every angle is non-obtuse. Further, since they are squares of positive side lengths, every angle is acute. So <math>a, b, c</math> are the side lengths of an <math>\textit{acute}</math> triangle.)
-WLOG, let <math>a < b < c</math>.
-For <math>a<4</math>, <math>a^2</math> is less than or equal to the gap between squares greater than <math>a^2</math>, so all such triangles are non-acute and fail.
-The next smallest case works:  <math>4^2 + 5^2 > 6^2</math> so <math>a, b, c = 4, 5, 6</math>.
-Using Heron's Formula, the minimum total surface area of the disphenoid is <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
-Bonus: by Heron’s Formula, the area of an <math>x-1, x, x+1</math> triangle is <math>x \sqrt{3 (x-2)(x+2)}/4 = x \sqrt{3 (x^2-2)}/4</math>.
-~babyhamster
-(Acute triangle observations and bonus formula by oinava)
-==Solution 3 (Volume formula)==
-Recall the formula for a disphenoid is <math>72V^2=(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)</math>, where <math>V</math> is the volume and <math>a,b,c</math> are the sides of each face. Because <math>72V^2</math> is a positive number, <math>(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)</math> must also be positive which is enough to rule out <math>2,3,4</math> and etc, leaving us with a <math>4-5-6</math> triangle for a surface area of <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>
--PaixiaoLover
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=23|num-a=25}}
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2024 AMC 12A Problems/Problem 24"

Revision as of 21:36, 20 March 2025

Problem

Solution

See also