Difference between revisions of "2023 AMC 12B Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | A rectangular box <math>P</math> has distinct edge lengths <math>a</math>, <math>b</math>, and <math>c</math>. The sum of the lengths of all <math>12</math> edges of <math>P</math> is <math>13</math>, the areas of all <math>6</math> faces of <math>P</math> is <math>\frac{11}{2}</math>, and the volume of <math>P</math> is <math>\frac{1}{2}</math>. What is the length of the longest interior diagonal connecting two vertices of <math>P</math>? | + | A rectangular box <math>\mathcal{P}</math> has distinct edge lengths <math>a</math>, <math>b</math>, and <math>c</math>. The sum of the lengths of all <math>12</math> edges of <math>\mathcal{P}</math> is <math>13</math>, the areas of all <math>6</math> faces of <math>\mathcal{P}</math> is <math>\frac{11}{2}</math>, and the volume of <math>\mathcal{P}</math> is <math>\frac{1}{2}</math>. What is the length of the longest interior diagonal connecting two vertices of <math>\mathcal{P}</math>? |
<math>\textbf{(A)}~2\qquad\textbf{(B)}~\frac{3}{8}\qquad\textbf{(C)}~\frac{9}{8}\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~\frac{3}{2}</math> | <math>\textbf{(A)}~2\qquad\textbf{(B)}~\frac{3}{8}\qquad\textbf{(C)}~\frac{9}{8}\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~\frac{3}{2}</math> | ||
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+ | ==Video Solution by MegaMath== | ||
+ | |||
+ | https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s | ||
==Solution 1 (algebraic manipulation)== | ==Solution 1 (algebraic manipulation)== | ||
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<cmath>2ab+2ac+2bc=\frac{11}{2}</cmath> | <cmath>2ab+2ac+2bc=\frac{11}{2}</cmath> | ||
<cmath>abc=\frac{1}{2}</cmath> | <cmath>abc=\frac{1}{2}</cmath> | ||
− | We also know that we want <math>\sqrt{a^2 + b^2 + c^2}</math> because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that <math>a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)</math>. We know that <math>a+b+c = \frac{13}{4}</math> and <math>2(ab+ac+bc)=\dfrac{11}2</math>, so <math>a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}</math>. So our answer is <math>\sqrt{\frac{81}{16}} = \boxed {\textbf{(D) \tfrac94}</math>. | + | We also know that we want <math>\sqrt{a^2 + b^2 + c^2}</math> because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that <math>a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)</math>. We know that <math>a+b+c = \frac{13}{4}</math> and <math>2(ab+ac+bc)=\dfrac{11}2</math>, so <math>a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}</math>. So our answer is <math>\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}</math>. |
Interestingly, we don't use the fact that the volume is <math>\frac{1}{2}</math>. | Interestingly, we don't use the fact that the volume is <math>\frac{1}{2}</math>. | ||
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~lprado | ~lprado | ||
− | ~minor edits and add-ons by Technodoggo, lucaswujc, and | + | ~minor edits and add-ons by Technodoggo, lucaswujc, andliu766, and BcMath |
− | ==Solution 2 ( | + | ==Solution 2 (Vieta's)== |
We use the equations from Solution 1 and manipulate it a little: | We use the equations from Solution 1 and manipulate it a little: | ||
<cmath>a+b+c = \frac{13}{4}</cmath> | <cmath>a+b+c = \frac{13}{4}</cmath> | ||
<cmath>ab+ac+bc=\frac{11}{4}</cmath> | <cmath>ab+ac+bc=\frac{11}{4}</cmath> | ||
<cmath>abc=\frac{1}{2}</cmath> | <cmath>abc=\frac{1}{2}</cmath> | ||
− | Notice how these are the equations for the vieta's formulas for a polynomial with roots of <math>a</math>, <math>b</math>, and <math>c</math>. Let's create that polynomial. It would be <math>x^3 - \frac{13}{4}x^2 + \frac{11}{4}x - \frac{1}{2}</math>. Multiplying each term by 4 to get rid of fractions, we get <math>4x^3 - 13x^2 + 11x - 2</math>. Notice how the coefficients add up to <math>0</math>. Whenever this happens, that means that <math>(x-1)</math> is a factor and that 1 is a root. After using synthetic division to divide <math>4x^3 - 13x^2 + 11x - 2</math> by <math>x-1</math>, we get <math>4x^2 - 9x + 2</math>. Factoring that, you get <math>(x-2)(4x-1)</math>. This means that this polynomial factors to <math>(x-1)(x-2)(4x-1)</math> and that the roots are <math>1</math>, <math>2</math>, and <math>1/4</math>. Since we're looking for <math>\sqrt{a^2 + b^2 + c^2}</math>, this is equal to <math>\sqrt{1^2 + 2^2 + \frac{1}{4}^2} = \sqrt{\frac{81}{16}} = \boxed {\textbf{(D) \tfrac94}</math> | + | Notice how these are the equations for the vieta's formulas for a polynomial with roots of <math>a</math>, <math>b</math>, and <math>c</math>. Let's create that polynomial. It would be <math>x^3 - \frac{13}{4}x^2 + \frac{11}{4}x - \frac{1}{2}</math>. Multiplying each term by 4 to get rid of fractions, we get <math>4x^3 - 13x^2 + 11x - 2</math>. Notice how the coefficients add up to <math>0</math>. Whenever this happens, that means that <math>(x-1)</math> is a factor and that 1 is a root. After using synthetic division to divide <math>4x^3 - 13x^2 + 11x - 2</math> by <math>x-1</math>, we get <math>4x^2 - 9x + 2</math>. Factoring that, you get <math>(x-2)(4x-1)</math>. This means that this polynomial factors to <math>(x-1)(x-2)(4x-1)</math> and that the roots are <math>1</math>, <math>2</math>, and <math>1/4</math>. Since we're looking for <math>\sqrt{a^2 + b^2 + c^2}</math>, this is equal to <math>\sqrt{1^2 + 2^2 + \frac{1}{4}^2} = \sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}</math> |
~lprado | ~lprado | ||
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==Solution 3 (Cheese Method)== | ==Solution 3 (Cheese Method)== | ||
− | Incorporating the solution above, we know <math>a+b+c</math> = <math> | + | Incorporating the solution above, we know <math>a+b+c</math> = <math>13/4</math> <math>\Rightarrow</math> <math>a+b+c > 3</math>. The side lengths are larger than <math>1</math> <math>\cdot</math> <math>1</math> <math>\cdot</math> <math>1</math> (a unit cube). The side length of the interior of a unit cube is <math>\sqrt{3}</math>, and we know that the side lengths are larger than <math>1</math> <math>\cdot</math> <math>1</math> <math>\cdot</math> <math>1</math>, so that means the diagonal has to be larger than <math>\sqrt{3}</math>, and the only answer choice larger than <math>\sqrt{3}</math> <math>\Rightarrow</math> <math>\boxed{\textbf{(D)}~\tfrac94}</math> |
~kabbybear | ~kabbybear | ||
− | Note that the real number <math>\sqrt{3}</math> is around <math>1.73</math>. Option <math>A</math> is also greater than <math>\sqrt{3}</math> meaning there are two options greater than <math>\sqrt{3}</math>. Option <math>A</math> is an integer so educationally guessing we arrive at answer <math>D</math> <math>\Rightarrow</math> <math>\boxed {\textbf{(D) \tfrac94}</math> | + | Note that the real number <math>\sqrt{3}</math> is around <math>1.73</math>. Option <math>A</math> is also greater than <math>\sqrt{3}</math> meaning there are two options greater than <math>\sqrt{3}</math>. Option <math>A</math> is an integer so educationally guessing we arrive at answer <math>D</math> <math>\Rightarrow</math> <math>\boxed{\textbf{(D)}~\tfrac94}</math> |
~atictacksh | ~atictacksh | ||
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==Video Solution 2 by SpreadTheMathLove== | ==Video Solution 2 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=lVkvcCmY9uM | https://www.youtube.com/watch?v=lVkvcCmY9uM | ||
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==Video Solution== | ==Video Solution== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by MegaMath== | ||
+ | |||
+ | https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s | ||
==See Also== | ==See Also== |
Latest revision as of 21:01, 9 November 2024
- The following problem is from both the 2023 AMC 10B #17 and 2023 AMC 12B #13, so both problems redirect to this page.
Contents
[hide]Problem
A rectangular box has distinct edge lengths , , and . The sum of the lengths of all edges of is , the areas of all faces of is , and the volume of is . What is the length of the longest interior diagonal connecting two vertices of ?
Video Solution by MegaMath
https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s
Solution 1 (algebraic manipulation)
We can create three equations using the given information. We also know that we want because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that . We know that and , so . So our answer is .
Interestingly, we don't use the fact that the volume is .
~lprado
~minor edits and add-ons by Technodoggo, lucaswujc, andliu766, and BcMath
Solution 2 (Vieta's)
We use the equations from Solution 1 and manipulate it a little: Notice how these are the equations for the vieta's formulas for a polynomial with roots of , , and . Let's create that polynomial. It would be . Multiplying each term by 4 to get rid of fractions, we get . Notice how the coefficients add up to . Whenever this happens, that means that is a factor and that 1 is a root. After using synthetic division to divide by , we get . Factoring that, you get . This means that this polynomial factors to and that the roots are , , and . Since we're looking for , this is equal to
~lprado
Solution 3 (Cheese Method)
Incorporating the solution above, we know = . 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
Note that the real number is around . Option is also greater than meaning there are two options greater than . Option is an integer so educationally guessing we arrive at answer
~atictacksh
Video Solution 1 by OmegaLearn
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=lVkvcCmY9uM
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by MegaMath
https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s
See Also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.