Difference between revisions of "2023 AMC 10B Problems/Problem 17"
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</cmath> | </cmath> | ||
Note the expansion of <cmath>\begin{align*} | Note the expansion of <cmath>\begin{align*} | ||
− | (a+b+c)^2 = a \cdot(a+b+c) + b \cdot(a+b+c) + c\cdot(a+b+c)\\ | + | & (a+b+c)^2 = a \cdot(a+b+c) + b \cdot(a+b+c) + c\cdot(a+b+c)\\ |
− | &=\ a^2 + ab + ac+ ab + b^2 + bc+ac+bc+c^2 = a^2+b^2+c^2 + 2ab + 2bc + 2ac) | + | &=\ a^2 + ab + ac+ ab + b^2 + bc+ac+bc+c^2 = \\ |
+ | &=\ a^2+b^2+c^2 + 2ab + 2bc + 2ac) | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> |
Revision as of 18:25, 15 November 2023
Contents
Problem
A rectangular box 𝒫 has distinct edge lengths 𝑎, 𝑏, and 𝑐. The sum of the lengths of all 12 edges of 𝒫 is 13, the sum of the areas of all 6 faces of 𝒫 is , and the volume of 𝒫 is . What is the length of the longest interior diagonal connecting two vertices of 𝒫 ?
Solution 1
Let and be the sides of the box, we get
The longest diagonal of prism will be the space diagonal, which can be found by summing the squares of all the sides and square rooting them.
Note the expansion of ~Technodoggo ~minor edits and add-ons by lucaswujc
Note
Interestingly, we don't use the fact that the volume is ~andliu766
Solution 2 (find side lengths)
Let be the edge lengths.
Then, you can notice that these look like results of Vieta's formula: Finding when this will give us the edge lengths. We can use RRT to find one of the roots: One is , dividing gives . The other 2 roots are
Then, once we find the 3 edges being and , we can plug in to the distance formula to get .
-HIA2020
Solution 3 (Cheese Method)
Incorporating the solution above, the side lengths are larger than (a unit cube). The side length of the interior of a unit cube is , and we know that the side lengths are larger than , so that means the diagonal has to be larger than , and the only answer choice larger than ~ kabbybear