Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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− | 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. <math>AC=\sqrt{49-x^2}</math>, <math>BC=\sqrt{25-x^2}</math>, <math>AB=\sqrt{49-x^2} - \sqrt{25-x^2} = 3</math>, <math>BD=\sqrt{49-x^2} + \sqrt{25-x^2}</math>. | + | 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>. |
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+ | 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> | ||
~isabelchen | ~isabelchen |
Revision as of 01:02, 23 October 2021
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, ,
~isabelchen
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 |