Difference between revisions of "2020 AMC 10B Problems/Problem 8"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Algebra): I found some flaws in this solution too. First, the equation was not solved correctly. Secondly, we should do casework which one is the right angle before we apply the Pythagorean Theorem.) |
MRENTHUSIASM (talk | contribs) (Polished the algebra solution.) |
||
Line 68: | Line 68: | ||
==Solution 2 (Algebra)== | ==Solution 2 (Algebra)== | ||
+ | Let the brackets denote areas. We are given that <cmath>[PQR]=\frac12\cdot PQ\cdot h_R=12.</cmath> Since <math>PQ=8,</math> it follows that <math>h_R=3.</math> | ||
+ | |||
+ | Without the loss of generality, let <math>P=(-4,0)</math> and <math>Q=(4,0).</math> We conclude that the <math>y</math>-coordinate of <math>R</math> must be <math>\pm3.</math> | ||
+ | |||
+ | We apply casework to the right angle of <math>\triangle PQR:</math> | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>\angle P=90^\circ.</math> <p> | ||
+ | The <math>x</math>-coordinate of <math>R</math> must be <math>-4,</math> so we have <math>R=(-4,\pm3).</math> <p> | ||
+ | In this case, there are <math>2</math> such locations for <math>R.</math><p></li> | ||
+ | <li><math>\angle Q=90^\circ.</math> <p> | ||
+ | The <math>x</math>-coordinate of <math>R</math> must be <math>4,</math> so we have <math>R=(4,\pm3).</math> <p> | ||
+ | In this case, there are <math>2</math> such locations for <math>R.</math><p></li> | ||
+ | <li><math>\angle R=90^\circ.</math> <p> | ||
+ | For <math>R=(x,3),</math> the Pythagorean Theorem <math>PR^2+QR^2=PQ^2</math> gives <cmath>\left[(x+4)^2+3^2\right]+\left[(x-4)^2+3^2\right]=8^2.</cmath> Solving this equation, we have <math>x=\pm\sqrt7,</math> or <math>R=\left(\pm\sqrt7,3\right).</math> <p> | ||
+ | For <math>R=(x,-3),</math> we have <math>R=\left(\pm\sqrt7,-3\right)</math> by a similar process. <p> | ||
+ | In this case, there are <math>4</math> such locations for <math>R.</math><p></li> | ||
+ | </ol> | ||
+ | Together, there are <math>2+2+4=\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math> | ||
~MRENTHUSIASM ~mewto | ~MRENTHUSIASM ~mewto |
Revision as of 07:44, 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 2 (Algebra)
Let the brackets denote areas. We are given that Since
it follows that
Without the loss of generality, let and
We conclude that the
-coordinate of
must be
We apply casework to the right angle of
The
-coordinate of
must be
so we have
In this case, there are
such locations for
The
-coordinate of
must be
so we have
In this case, there are
such locations for
For
the Pythagorean Theorem
gives
Solving this equation, we have
or
For
we have
by a similar process.
In this case, there are
such locations for
Together, there are such locations for
~MRENTHUSIASM ~mewto
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.