Difference between revisions of "2022 AMC 10A Problems/Problem 13"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 1 ( | + | ==Solution 1 (Angle Bisector Theorem and Similar Triangles)== |
− | |||
− | + | Suppose that <math>\overline{BD}</math> intersects <math>\overline{AP}</math> and <math>\overline{AC}</math> at <math>X</math> and <math>Y,</math> respectively. By Angle-Side-Angle, we conclude that <math>\triangle ABX\cong\triangle AYX.</math> | |
− | + | Let <math>AB=AY=2x.</math> By the Angle Bisector Theorem, we have <math>AC=3x,</math> or <math>YC=x.</math> | |
− | + | By alternate interior angles, we get <math>\angle YAD=\angle YCB</math> and <math>\angle YDA=\angle YBC.</math> Note that <math>\triangle ADY \sim \triangle CBY</math> by the Angle-Angle Similarity, with the ratio of similitude <math>\frac{AY}{CY}=2.</math> It follows that <math>AD=2CB=2(BP+PC)=\boxed{\textbf{(C) } 10}.</math> | |
− | ~ | + | ~MRENTHUSIASM |
− | ==Solution 2 ( | + | ==Solution 2 (Auxiliary Lines)== |
+ | Let the intersection of <math>AC</math> and <math>BD</math> be <math>M</math>, and the intersection of <math>AP</math> and <math>BD</math> be <math>N</math>. Draw a line from <math>M</math> to <math>BC</math>, and label the point of intersection <math>O</math>. | ||
− | + | By adding this extra line, we now have many pairs of similar triangles. We have <math>\triangle BPN \sim \triangle BOM</math>, with a ratio of <math>2</math>, so <math>BO = 4</math> and <math>OC = 1</math>. We also have <math>\triangle COM \sim \triangle CAP</math> with ratio <math>3</math>. Additionally, <math>\triangle BPN \sim \triangle ADN</math> (with an unknown ratio). It is also true that <math>\triangle BAN \cong \triangle MAN</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</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 |
− | == Solution 3 ( | + | == Solution 3 (Slopes) == |
Let point <math>B</math> be the origin, with <math>C</math> being on the positive <math>x</math>-axis and <math>A</math> being in the first quadrant. | Let point <math>B</math> be the origin, with <math>C</math> being on the positive <math>x</math>-axis and <math>A</math> being in the first quadrant. | ||
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Let the perpendicular from <math>A</math> to <math>BC</math> be <math>AM</math>. | Let the perpendicular from <math>A</math> to <math>BC</math> be <math>AM</math>. | ||
− | + | Using Heron's formula, <cmath>[ABC] = \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} = \frac{15}{4}\sqrt{7}.</cmath> | |
Hence, <cmath>AM = \frac{\frac{15}{4}\sqrt{7}}{\frac{5}{2}} = \frac{3}{2}\sqrt{7}.</cmath> | Hence, <cmath>AM = \frac{\frac{15}{4}\sqrt{7}}{\frac{5}{2}} = \frac{3}{2}\sqrt{7}.</cmath> | ||
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~[[User:Bxiao31415|Bxiao31415]] | ~[[User:Bxiao31415|Bxiao31415]] | ||
− | == Video Solution 1 == | + | ==Video Solution 1 == |
+ | https://youtu.be/m1-7E8T_i_E | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | == Video Solution 2 == | ||
https://youtu.be/_0_EGdkhOFg | https://youtu.be/_0_EGdkhOFg | ||
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- Whiz | - Whiz | ||
− | == Video Solution | + | == Video Solution 3 == |
https://youtu.be/77JIN0iVizA | https://youtu.be/77JIN0iVizA | ||
− | + | == Video Solution 4 == | |
+ | |||
+ | https://youtu.be/G8NRcVxSdz0 | ||
− | ==Video Solution | + | ==Video Solution 5 by SpreadTheMathLove== |
− | https:// | + | |
+ | https://www.youtube.com/watch?v=nhlpSATltRU | ||
+ | |||
+ | ~Ismail.maths93 | ||
− | + | == Video Solution 6 by Lucas637 == | |
+ | https://www.youtube.com/watch?v=R1CtcZ2pWVk | ||
== See Also == | == See Also == |
Latest revision as of 08:04, 19 May 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Angle Bisector Theorem and Similar Triangles)
- 4 Solution 2 (Auxiliary Lines)
- 5 Solution 3 (Slopes)
- 6 Solution 4 (Assumption)
- 7 Video Solution 1
- 8 Video Solution 2
- 9 Video Solution 3
- 10 Video Solution 4
- 11 Video Solution 5 by SpreadTheMathLove
- 12 Video Solution 6 by Lucas637
- 13 See Also
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 (Angle Bisector Theorem and Similar Triangles)
Suppose that intersects
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 2 (Auxiliary Lines)
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 pairs of similar triangles. 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 3 (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
.
Using Heron's formula,
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
~Education, the Study of Everything
Video Solution 2
- Whiz
Video Solution 3
Video Solution 4
Video Solution 5 by SpreadTheMathLove
https://www.youtube.com/watch?v=nhlpSATltRU
~Ismail.maths93
Video Solution 6 by Lucas637
https://www.youtube.com/watch?v=R1CtcZ2pWVk
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.