Difference between revisions of "2020 AMC 10B Problems/Problem 8"
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We construct a circle with diameter <math>\overline{PQ}.</math> All such locations for <math>R</math> are shown below: | We construct a circle with diameter <math>\overline{PQ}.</math> All such locations for <math>R</math> are shown below: | ||
+ | |||
<asy> | <asy> | ||
/* Made by MRENTHUSIASM */ | /* Made by MRENTHUSIASM */ | ||
Line 45: | Line 46: | ||
dot(I1,linewidth(4)); | dot(I1,linewidth(4)); | ||
dot(I2,linewidth(4)); | dot(I2,linewidth(4)); | ||
− | + | Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); | |
− | Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(white)); | + | Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); |
− | Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(white)); | ||
draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15)); | draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15)); | ||
draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); | draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); | ||
draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); | draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); | ||
</asy> | </asy> | ||
+ | |||
We apply casework to the right angle of <math>\triangle PQR:</math> | We apply casework to the right angle of <math>\triangle PQR:</math> | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> |
Revision as of 11:17, 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.