Difference between revisions of "2018 AMC 10A Problems/Problem 10"
(→Solutions) |
(→Video Solution) |
||
Line 76: | Line 76: | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zQG70XKAdeA | ||
+ | ~ North America Math Contest Go Go Go | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:04, 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 |