Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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===Solution 6 (Symmetric Substitution)=== | ===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{8}</math>. | + | 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)=== | ===Solution 7 (Difference of Squares)=== |
Revision as of 09:22, 28 September 2022
Contents
[hide]Problem
Suppose that real number satisfies What is the value of ?
Solutions
Solution 1
In order to eliminate 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 it as . Given the in the problem, we can divide .
-aze.10
Solution 6 (Symmetric Substitution)
Since , let . Then we have . Squaring both sides gives us . Isolating the term with the square root, and squaring again, we get . Then .
Solution 7 (Difference of Squares)
Let and . Then by difference of squares:
.
We can simplify this expression to get our answer. and from the given statement, . Now we have:
.
Hence, so our answer is .
~BakedPotato66
Solution 8 (Analytic Geometry)
The problem can be represented by the above diagram. The large circle with center has a radius of 7, the small circle with center has a radius of 5. Point 's X coordinate is . , , , .
By Power of a Point, ,
Solution 9 (Pythagorean Theorem)
Notice that and 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, and are the hypotenuses of the two triangles, and is the one leg from each of the triangles. So, is the other leg of the 1st one, and is the other leg of the 2nd one.
For convenience, we name the other leg of the 1st triangle , and the other leg of the 2nd one . Using the Pythagorean Theorem, we set up 2 equations.
Subtracting the two equations and canceling out , we have , which simplifies to .
We already know that (or is equal to , so plugging it in, we have , and dividing by gives
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
Video Solution 4
~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 |