Difference between revisions of "2018 AMC 10A Problems/Problem 10"

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\textbf{(E) }12\qquad
 
\textbf{(E) }12\qquad
 
</math>
 
</math>
 
==Solutions==
 
 
===Solution 1===
 
In order to eliminate 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 6 (Symmetric Substitution)===
 
Since <math>\frac{25+49}{2}=37</math>, let <math>37-x^2 = y</math>. Then we have <math>\sqrt{y+12}-\sqrt{y-12}=3</math>. Squaring both sides gives us <math>2y-2\sqrt{y^2-144}=9</math>. Isolating the term with the square root, and squaring again, we get <math>4y^2-36y+81=4y^2-576 \implies y=\frac{73}{4}</math>. Then <math>\sqrt{y+12}+\sqrt{y-12} = \sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}} = \frac{16}{2}=\boxed{\textbf{(A)}\ 8}</math>.
 
 
===Solution 7 (Difference of Squares)===
 
Let <math>\sqrt{49-x^2}=a</math> and <math>\sqrt{25-x^2}=b</math>. Then by difference of squares:
 
 
<math>(a+b)(a-b)=a^2-b^2</math>.
 
 
We can simplify this expression to get our answer. <math>a^2-b^2=(49-x^2)-(25-x^2)=24</math> and from the given statement, <math>a-b=3</math>. Now we have:
 
 
<math>(a+b)(3)=24</math>.
 
 
Hence, <math>a+b=\sqrt{49-x^2}-\sqrt{25-x^2}=8</math> so our answer is <math>\boxed{\textbf{(A) } 8}</math>.
 
 
~BakedPotato66
 
 
===Solution 8 (Analytic Geometry)===
 
 
[[File:2018 AMC10 A P10.PNG|500px]]
 
 
The problem can be represented by the above diagram. The large circle with center <math>O</math> has a radius of 7, the small circle with center <math>O</math> has a radius of 5. Point <math>C</math>'s X coordinate is <math>x</math>. <math>AC=CD=\sqrt{49-x^2}</math>, <math>BC=\sqrt{25-x^2}</math>, <math>AB=AC-BC=\sqrt{49-x^2} - \sqrt{25-x^2} = 3</math>, <math>BD=CD+BC=\sqrt{49-x^2} + \sqrt{25-x^2}</math>.
 
 
By Power of a Point, <math>AB \cdot BD=BE \cdot BF=(7-5) \cdot (7+5)=24</math>, <math>BD=\boxed{\textbf{(A) } 8}</math>
 
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
 
===Solution 9 (Pythagorean Theorem)===
 
 
Notice that <math>\sqrt{49-x^2} = \sqrt{7^2-x^2}</math> and <math>\sqrt{25-x^2} = \sqrt{5^2-x^2}</math> This is also the equation of finding a leg of a right triangle given the hypotenuse and the other leg using the [[Pythagorean Theorem]].
 
 
Now, <math>7</math> and <math>5</math> are the hypotenuses of the two triangles, and <math>x</math> is the one leg from each of the triangles. So, <math>\sqrt{7^2-x^2}</math> is the other leg of the 1st one, and <math>\sqrt{5^2-x^2}</math> is the other leg of the 2nd one.
 
 
For convenience, we name the other leg of the 1st triangle <math>a</math> (the one that's not <math>5</math> or <math>x</math>), and the other leg of the 2nd one <math>b</math> (the one that's not <math>7</math> or <math>x</math>). Using the Pythagorean Theorem, we set up 2 equations.
 
<cmath>\begin{align*}
 
a^2 + x^2 &= 7^2 \
 
b^2 + x^2 &= 5^2
 
\end{align*}</cmath>
 
 
Subtracting the two equations and canceling out <math>x^2</math>, we have <math>a^2 - b^2 = 49-25</math>, which simplifies to <math>(a-b)(a+b)=24</math>.
 
 
We already know that <math>a-b</math> (or <math>\sqrt{49-x^2}-\sqrt{25-x^2}</math>) is equal to <math>3</math>, so plugging it in, we have <math>3(a+b)=24</math>, and dividing by <math>3</math> gives <math>a+b = \boxed{\textbf{(A)}\ 8}</math>
 
 
~MrThinker
 
  
 
==Solution 10 (Solution 1 but alternate)==
 
==Solution 10 (Solution 1 but alternate)==

Revision as of 18:44, 25 March 2024

Problem

Suppose that real number $x$ satisfies \[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?

$\textbf{(A) }8\qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9\qquad \textbf{(D) }2\sqrt{10}+4\qquad \textbf{(E) }12\qquad$

Solution 10 (Solution 1 but alternate)

We let $a=\sqrt{49-x^2}+\sqrt{25-x^2}$; in other words, we want to find $a$. We know that $a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.$ Thus, $a=8.$

~Technodoggo

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/P-atxiiTw2I

~Education, the Study of Everything



Video Solutions

Video Solution 1

https://youtu.be/ba6w1OhXqOQ?t=1403

~ pi_is_3.14

Video Solution 2

https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go

Video Solution 3

https://youtu.be/ZiZVIMmo260

Video Solution 4

https://youtu.be/5cA87rbzFdw

~savannahsolver

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 10 Problems and Solutions