Difference between revisions of "2023 AMC 10B Problems/Problem 17"
Technodoggo (talk | contribs) (→Solution) |
Technodoggo (talk | contribs) (→Solution) |
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label("c",midpoint(AA--CC),S); | label("c",midpoint(AA--CC),S); | ||
</asy> | </asy> | ||
+ | Let <math>a,b,</math> and <math>c</math> be the sides of the box, we get | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 4(a+b+c) &= 13\\ | ||
+ | 2(ab+bc+ca) &= \dfrac{11}{2}\\ | ||
+ | abc &= \dfrac{1}{2} | ||
+ | \end{align*}</cmath> | ||
+ | |||
The diagonal of the box is | 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{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{13}{4})^2-\dfrac{11}{2}}\\ | ||
+ | &=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\ | ||
+ | &=\sqrt{\dfrac{81}{16}}\\ | ||
&=\dfrac{9}{4} | &=\dfrac{9}{4} | ||
\end{align*} | \end{align*} |
Revision as of 15:32, 15 November 2023
Solution
Let and be the sides of the box, we get
The diagonal of the box is
~Technodoggo