Difference between revisions of "2023 AMC 10B Problems/Problem 17"

Revision as of 16:32, 15 November 2023

Solution

$[asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); draw(D--AA,dashed); draw(A--B); draw(A--C); draw(B--D); draw(C--D); draw(A--AA); draw(B--BB); draw(C--CC); draw(D--DD); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB); draw(AA--CC); draw(BB--DD); draw(CC--DD); label("a",midpoint(D--DD),E); label("b",midpoint(CC--DD),E); label("c",midpoint(AA--CC),S); [/asy]$ Let $a,b,$ and $c$ be the sides of the box, we get

$\begin{align*} 4(a+b+c) &= 13\\ 2(ab+bc+ca) &= \dfrac{11}{2}\\ abc &= \dfrac{1}{2} \end{align*}$

The diagonal of the box is

$\begin{align*} \sqrt{a^2+b^2+c^2}&=\sqrt{(a+b+c)^2-2(ab+bc+ca)}\\ &=\sqrt{(\dfrac{13}{4})^2-\dfrac{11}{2}}\\ &=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\ &=\sqrt{\dfrac{81}{16}}\\ &=\dfrac{9}{4} \end{align*}$

~Technodoggo

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Difference between revisions of "2023 AMC 10B Problems/Problem 17"

Revision as of 16:32, 15 November 2023

Solution

@@ Line 48: / Line 48: @@
 label("c",midpoint(AA--CC),S);
 </asy>
+Let <math>a,b,</math> and <math>c</math> be the sides of the box, we get
+<cmath>\begin{align*}
+(a+b+c) &= 13\\
+(ab+bc+ca) &= \dfrac{11}{2}\\
+abc &= \dfrac{1}{2}
+\end{align*}</cmath>
 The diagonal of the box is
@@ Line 54: / Line 62: @@
     \sqrt{a^2+b^2+c^2}&=\sqrt{(a+b+c)^2-2(ab+bc+ca)}\\
 &=\sqrt{(\dfrac{13}{4})^2-\dfrac{11}{2}}\\
+&=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\
+&=\sqrt{\dfrac{81}{16}}\\
 &=\dfrac{9}{4}
 \end{align*}