Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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− | Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath> | + | ==Problem== |
+ | |||
+ | Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath>What is the value of <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>? | ||
<math> | <math> | ||
− | \textbf{(A) }8 \qquad | + | \textbf{(A) }8\qquad |
\textbf{(B) }\sqrt{33}+8\qquad | \textbf{(B) }\sqrt{33}+8\qquad | ||
− | \textbf{(C) }9 \qquad | + | \textbf{(C) }9\qquad |
− | \textbf{(D) }2\sqrt{10}+4 \qquad | + | \textbf{(D) }2\sqrt{10}+4\qquad |
− | \textbf{(E) }12 \qquad | + | \textbf{(E) }12\qquad |
</math> | </math> | ||
− | == Solution 1== | + | ==Solutions== |
+ | |||
+ | ===Solution 1=== | ||
+ | In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The <math>x^2</math> terms cancel nicely. <math>(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math> | ||
+ | |||
+ | Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3, (\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}</math>. - cookiemonster2004 | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Let <math>u=\sqrt{49-x^2}</math>, and let <math>v=\sqrt{25-x^2}</math>. Then <math>v=\sqrt{u^2-24}</math>. Substituting, we get <math>u-\sqrt{u^2-24}=3</math>. Rearranging, we get <math>u-3=\sqrt{u^2-24}</math>. Squaring both sides and solving, we get <math>u=\frac{11}{2}</math> and <math>v=\frac{11}{2}-3=\frac{5}{2}</math>. Adding, we get that the answer is <math>\boxed{\textbf{(A) } 8}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | Put the equations to one side. <math>\sqrt{49-x^2}-\sqrt{25-x^2}=3</math> can be changed into <math>\sqrt{49-x^2}=\sqrt{25-x^2}+3</math>. | ||
+ | |||
+ | We can square both sides, getting us <math>49-x^2=(25-x^2)+(3^2)+ 2\cdot 3 \cdot \sqrt{25-x^2}.</math> | ||
+ | |||
+ | That simplifies out to <math>15=6 \sqrt{25-x^2}.</math> Dividing both sides by <math>6</math> gets us <math>\frac{5}{2}=\sqrt{25-x^2}</math>. | ||
+ | |||
+ | Following that, we can square both sides again, resulting in the equation <math>\frac{25}{4}=25-x^2</math>. Simplifying that, we get <math>x^2 = \frac{75}{4}</math>. | ||
+ | |||
+ | Substituting into the equation <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>, we get <math>\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}</math>. Immediately, we simplify into <math>\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}</math>. The two numbers inside the square roots are simplified to be <math>\frac{11}{2}</math> and <math>\frac{5}{2}</math>, so you add them up: <math>\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}</math>. | ||
+ | |||
+ | ~kevinmathz | ||
+ | |||
+ | ===Solution 4 (Geometric Interpretation)=== | ||
+ | |||
+ | Draw a right triangle <math>ABC</math> with a hypotenuse <math>AC</math> of length <math>7</math> and leg <math>AB</math> of length <math>x</math>. Draw <math>D</math> on <math>BC</math> such that <math>AD=5</math>. Note that <math>BC=\sqrt{49-x^2}</math> and <math>BD=\sqrt{25-x^2}</math>. Thus, from the given equation, <math>BC-BD=DC=3</math>. Using Law of Cosines on triangle <math>ADC</math>, we see that <math>\angle{ADC}=120^{\circ}</math> so <math>\angle{ADB}=60^{\circ}</math>. Since <math>ADB</math> is a <math>30-60-90</math> triangle, <math>\sqrt{25-x^2}=BD=\frac{5}{2}</math> and <math>\sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}</math>. Finally, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}</math>. | ||
+ | <asy> | ||
+ | var s = sqrt(3); | ||
+ | pair A = (-5*s/2, 0); | ||
+ | pair B = (0,0); | ||
+ | pair C = (0,5.5); | ||
+ | pair D = (0,2.5); | ||
+ | |||
+ | draw(A--B--C--A--D); | ||
+ | rightanglemark(A, B, D); | ||
+ | label("A", A, SW); | ||
+ | label("B", B, SE); | ||
+ | label("C", C, NE); | ||
+ | label("D", D, E); | ||
+ | label("7", (-5*s/4, 5.5/2), NW); | ||
+ | label("120$^\circ$", D, NW); | ||
+ | label("60$^\circ$", (0,2), SW); | ||
+ | label("$x$", 0.5*A, S); | ||
+ | draw(rightanglemark(A, B, C)); | ||
+ | |||
+ | draw(anglemark(A, D, B)); | ||
+ | markscalefactor = 0.04; | ||
+ | draw(anglemark(C, D, A)); | ||
− | + | label("$\frac{5}{2}$", (0,1.25), E); | |
+ | label("3", (0,4), E); | ||
+ | label("5", (-5*s/4, 5/4), N); | ||
+ | </asy> | ||
− | + | ===Solution 5 (No Square Roots, Fastest)=== | |
+ | We notice that the two expressions are conjugates, and therefore we can write them in a "difference-of-squares" format. | ||
+ | Namely, we can write is as <math>((49-x^2) - (25 - x^2)) = (49 - x^2 - 25 + x^2) = 24</math>. Given the <math>3</math> in the problem, we can divide <math>24 / 3 = \boxed{\textbf{(A) } 8}</math>. | ||
− | + | -aze.10 | |
− | |||
− | |||
− | == See Also == | + | == Video Solution == |
+ | https://youtu.be/ba6w1OhXqOQ?t=1403 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zQG70XKAdeA | ||
+ | ~ North America Math Contest Go Go Go | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | https://youtu.be/5cA87rbzFdw | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
{{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | ||
− | + | ||
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 16:11, 31 January 2021
Contents
Problem
Suppose that real number satisfies What is the value of ?
Solutions
Solution 1
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The terms cancel nicely.
Given that . - cookiemonster2004
Solution 2
Let , and let . Then . Substituting, we get . Rearranging, we get . Squaring both sides and solving, we get and . Adding, we get that the answer is .
Solution 3
Put the equations to one side. can be changed into .
We can square both sides, getting us
That simplifies out to Dividing both sides by gets us .
Following that, we can square both sides again, resulting in the equation . Simplifying that, we get .
Substituting into the equation , we get . Immediately, we simplify into . The two numbers inside the square roots are simplified to be and , so you add them up: .
~kevinmathz
Solution 4 (Geometric Interpretation)
Draw a right triangle with a hypotenuse of length and leg of length . Draw on such that . Note that and . Thus, from the given equation, . Using Law of Cosines on triangle , we see that so . Since is a triangle, and . Finally, .
Solution 5 (No Square Roots, Fastest)
We notice that the two expressions are conjugates, and therefore we can write them in a "difference-of-squares" format. Namely, we can write is as . Given the in the problem, we can divide .
-aze.10
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1403
~ pi_is_3.14
Video Solution
https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go
Video Solution
~savannahsolver
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |