Difference between revisions of "2020 AMC 10B Problems/Problem 8"
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==Solution 4 (Algebra)== | ==Solution 4 (Algebra)== | ||
− | Let a and b be the distances of R from P and Q, respectively. By the Pythagorean Theorem, we have < | + | Let <math>a</math> and <math>b</math> be the distances of <math>R</math> from <math>P</math> and <math>Q,</math> respectively. |
+ | |||
+ | By the Pythagorean Theorem, we have <cmath>a^2 + b^2 = 64.</cmath> | ||
+ | Since the area of this triangle is <math>12,</math> we get <math>a \cdot b = 12 \cdot 2 = 24.</math> Thus, <math>b = \frac{24}{a}.</math> Now substitute this into the other equation to get <cmath>a^2 + \left(\frac{24}{a}\right)^2 = 64.</cmath> | ||
+ | Multiplying by <math>a^2</math> on both sides, we get <cmath>\text{There is flaw here. I will fix it tomorrow. Bedtime for me. }a^4 + 24 = 64a^2.</cmath> | ||
+ | Now let <math>y = a^2.</math> Substituting and rearranging, we get <cmath>y^2 - 64y + 24 = 0.</cmath> | ||
+ | We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are <math>y = 32 \pm 10\sqrt{10}.</math> Now substitute back <math>y = a^2</math> to get <math>a = \pm \sqrt{32 \pm 10\sqrt{10}}.</math> All <math>4</math> of these solutions are rational and will work. But our answer is actually <math>4\cdot2 = 8</math> as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ. | ||
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~mewto | ~mewto | ||
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+ | \textbf{(D)}\ 8 | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:37, 30 September 2021
Problem
Points and
lie in a plane with
. How many locations for point
in this plane are there such that the triangle with vertices
,
, and
is a right triangle with area
square units?
Solution 1 (Geometry)
Let the brackets denote areas. We are given that Since
it follows that
We construct a circle with diameter All such locations for
are shown below:
We apply casework to the right angle of
- If
then
by the tangent.
- If
then
by the tangent.
- If
then
by the Inscribed Angle Theorem.
Together, there are such locations for
Remarks
- The reflections of
about
are
respectively.
- The reflections of
about the perpendicular bisector of
are
respectively.
~MRENTHUSIASM
Solution 4 (Algebra)
Let and
be the distances of
from
and
respectively.
By the Pythagorean Theorem, we have
Since the area of this triangle is
we get
Thus,
Now substitute this into the other equation to get
Multiplying by
on both sides, we get
Now let
Substituting and rearranging, we get
We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are
Now substitute back
to get
All
of these solutions are rational and will work. But our answer is actually
as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ.
~mewto
\textbf{(D)}\ 8
Video Solution
~IceMatrix
~savannahsolver
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.