Difference between revisions of "2022 AMC 10A Problems/Problem 13"
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Suppose the area of <math>\triangle COM</math> is <math>x</math>. Then, <math>[\triangle CAP] = 9x</math>. Because <math>\triangle CAP</math> and <math>\triangle BAP</math> share the same height and have a base ratio of <math>3:2</math>, <math>[\triangle BAP] = 6x</math>. Because <math>\triangle BOM</math> and <math>\triangle COM</math> share the same height and have a base ratio of <math>4:1</math>, <math>[\triangle BOM] = 4x</math>, <math>[\triangle BPN] = x</math>, and thus <math>[OMNP] = 4x - x = 3x</math>. Thus, <math>[\triangle MAN] = [\triangle BAN] = 5x</math>. | Suppose the area of <math>\triangle COM</math> is <math>x</math>. Then, <math>[\triangle CAP] = 9x</math>. Because <math>\triangle CAP</math> and <math>\triangle BAP</math> share the same height and have a base ratio of <math>3:2</math>, <math>[\triangle BAP] = 6x</math>. Because <math>\triangle BOM</math> and <math>\triangle COM</math> share the same height and have a base ratio of <math>4:1</math>, <math>[\triangle BOM] = 4x</math>, <math>[\triangle BPN] = x</math>, and thus <math>[OMNP] = 4x - x = 3x</math>. Thus, <math>[\triangle MAN] = [\triangle BAN] = 5x</math>. | ||
− | Finally, we have <math>\frac{[\triangle BAN]}{[\triangle BPN]} = \frac{5x}{x} = 5</math>, and because these triangles share the same height <math>\frac{AN}{PN} = 5. Notice that these side lengths are corresponding side lengths of the similar triangles < | + | Finally, we have <math>\frac{[\triangle BAN]}{[\triangle BPN]} = \frac{5x}{x} = 5</math>, and because these triangles share the same height <math>\frac{AN}{PN} = 5</math>. Notice that these side lengths are corresponding side lengths of the similar triangles <math>BPN</math> and <math>ADN</math>. This means that <math>AD = 5\cdot BP = \boxed{\textbf{(C) } 10}</math>. |
~mathboy100 | ~mathboy100 |
Revision as of 02:26, 26 November 2022
Contents
Problem
Let be a scalene triangle. Point
lies on
so that
bisects
The line through
perpendicular to
intersects the line through
parallel to
at point
Suppose
and
What is
Diagram
~MRENTHUSIASM
Solution 1 (The extra line)
Let the intersection of and
be
, and the intersection of
and
be
. Draw a line from
to
, and label the point of intersection
.
By adding this extra line, we now have many similarities. We have , with a ratio of
, so
and
. We also have
with ratio
. Additionally,
(with an unknown ratio). It is also true that
.
Suppose the area of is
. Then,
. Because
and
share the same height and have a base ratio of
,
. Because
and
share the same height and have a base ratio of
,
,
, and thus
. Thus,
.
Finally, we have , and because these triangles share the same height
. Notice that these side lengths are corresponding side lengths of the similar triangles
and
. This means that
.
~mathboy100
Solution 2 (Generalization)
Suppose that intersect
and
at
and
respectively. By Angle-Side-Angle, we conclude that
Let By the Angle Bisector Theorem, we have
or
By alternate interior angles, we get and
Note that
by the Angle-Angle Similarity, with the ratio of similitude
It follows that
~MRENTHUSIASM
Solution 3 (Coord geo, slopes)
Let point be the origin, with
being on the positive
-axis and
being in the first quadrant.
By the Angle Bisector Theorem, . Thus, assume that
, and
.
Let the perpendicular from to
be
.
We have
Hence,
Next, we have
The slope of line is thus
Therefore, since the slopes of perpendicular lines have a product of , the slope of line
is
. This means that we can solve for the coordinates of
:
We also know that the coordinates of are
, because
and
.
Since the -coordinates of
and
are the same, and their
-coordinates differ by
, the distance between them is
. Our answer is
~mathboy100
Solution 4 (Assumption)
Since there is only one possible value of
, we assume
. By the angle bisector theorem,
, so
and
. Now observe that
. Let the intersection of
and
be
. Then
. Consequently,
and therefore
, so
, and we're done!
Video Solution 1
- Whiz
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution (Quick and Simple)
~Education, the Study of Everything
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.