Difference between revisions of "2020 AMC 10B Problems/Problem 8"
MRENTHUSIASM (talk | contribs) (Polished the algebra solution.) |
MRENTHUSIASM (talk | contribs) (→Solution 2 (Algebra)) |
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<li><math>\angle P=90^\circ.</math> <p> | <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> | 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> | + | <b>In this case, there are <math>\boldsymbol{2}</math> such locations for <math>\boldsymbol{R.}</math></b> <p></li> |
<li><math>\angle Q=90^\circ.</math> <p> | <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> | 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> | + | <b>In this case, there are <math>\boldsymbol{2}</math> such locations for <math>\boldsymbol{R.}</math></b> <p></li> |
<li><math>\angle R=90^\circ.</math> <p> | <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> 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> | 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> | + | <b>In this case, there are <math>\boldsymbol{4}</math> such locations for <math>\boldsymbol{R.}</math></b> <p></li> |
</ol> | </ol> | ||
Together, there are <math>2+2+4=\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math> | Together, there are <math>2+2+4=\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math> |
Revision as of 08:01, 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.