Difference between revisions of "2023 AMC 12B Problems/Problem 13"

Revision as of 19:43, 15 November 2023

The following problem is from both the 2023 AMC 10B #17 and 2023 AMC 12B #13, so both problems redirect to this page.

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 areas of all 6 faces of P is $\frac{11}{2}$ , and the volume of P is $\frac{1}{2}$ . What is the length of the longest interior diagonal connecting two vertices of P?

Solution 1 (algebraic manipulation)

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

We can create three equations using the given information. $4a+4b+4c = 13$ $2ab+2ac+2bc=\frac{11}{2}$ $abc=\frac{1}{2}$ We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We know that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2ab - 2ac - 2bc$ . $a+b+c = \frac{13}{4}$ . So $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$ . So our answer is $\sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}$ .

Interestingly, we don't use the fact that the volume is $\frac{1}{2}$ . ~andliu766

~lprado ~Technodoggo ~minor edits and add-ons by lucaswujc

Solution 2 (factoring a polynomial)

We use the equations from Solution 1 and manipulate it a little: $a+b+c = \frac{13}{4}$ $ab+ac+bc=\frac{11}{4}$ $abc=\frac{1}{2}$ Notice how these are the equations for the vieta's formulas for a polynomial with roots of $a$ , $b$ , and $c$ . Let's create that polynomial. It would be $x^3 - \frac{13}{4}x^2 + \frac{11}{4}x - \frac{1}{2}$ . Multiplying each term by 4 to get rid of fractions, we get $4x^3 - 13x^2 + 11x - 2$ . Notice how the coefficients add up to $0$ . Whenever this happens, that means that $(x-1)$ is a factor and that 1 is a root. After using synthetic division to divide $4x^3 - 13x^2 + 11x - 2$ by $x-1$ , we get $4x^2 - 9x + 2$ . Factoring that, you get $(x-2)(4x-1)$ . This means that this polynomial factors to $(x-1)(x-2)(4x-1)$ and that the roots are $1$ , $2$ , and $1/4$ . Since we're looking for $\sqrt{a^2 + b^2 + c^2}$ , this is equal to $\sqrt{1^2 + 2^2 + \frac{1}{4}^2} = \sqrt{\frac{81}{16}} = \boxed{\frac{9}{4}}$

~lprado

Solution 3 (find side lengths)

Let $a,b,c$ be the edge lengths. $4(a+b+c)=13, a+b+c=13/4$ $2(ab+bc+ac)=11/2, ab+bc+ac=11/4$ $abc=1/2$

Then, you can notice that these look like results of Vieta's formula: $(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc = x^3-13/4x^2+11/4x-1/2$ Finding when this $= 0$ will give us the edge lengths. We can use RRT to find one of the roots: One is $x=1$ , dividing gives $x^2-9/4x+1/2$ . The other 2 roots are $2,1/4$

Then, once we find the 3 edges being $a=1,b=2,$ and $c=1/4$ , we can plug in to the distance formula to get $9/4$ .

-HIA2020

Solution 4 (Cheese Method)

Incorporating the solution above, the side lengths are larger than $1$ $\cdot$ $1$ $\cdot$ $1$ (a unit cube). The side length of the interior of a unit cube is $\sqrt{3}$ , and we know that the side lengths are larger than $1$ $\cdot$ $1$ $\cdot$ $1$ , so that means the diagonal has to be larger than $\sqrt{3}$ , and the only answer choice larger than $\sqrt{3}$ $\Rightarrow$ $\boxed {\textbf{(D) 9/4}}$

~kabbybear

@@ Line 101: / Line 101: @@
 ==See Also==
-{{AMC12 box|year=2023|ab=B|num-b=16|num-a=18}}
+{{AMC10 box|year=2023|ab=B|num-b=16|num-a=18}}
 {{AMC12 box|year=2023|ab=B|num-b=12|num-a=14}}
 {{MAA Notice}}

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Preceded by Problem 16	Followed by Problem 18
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Preceded by Problem 12	Followed by Problem 14
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