Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2009 AIME I Problems/Problem 5: Difference between revisions

← Older editNewer edit →
Motoe123 (talk | contribs)
33 edits
Kubluck (talk | contribs)
307 edits
Line 5: Line 5:


== Solution ==
== Solution ==
Sorry, I fail to get the diagram up here, someone help me.
Sorry, I failed to get the diagram up here, someone help me.


Since <math>K</math> is the midpoint of <math>\overline{PM}, \overline{AC}</math>.
Since <math>K</math> is the midpoint of <math>\overline{PM}, \overline{AC}</math>.
Line 19: Line 19:
That makes <math>\bigtriangleup{AMB}</math> congruent to <math>\bigtriangleup{LPB}</math>
That makes <math>\bigtriangleup{AMB}</math> congruent to <math>\bigtriangleup{LPB}</math>


Thus, <math>\frac {AM}{LP}=\frac {AB}{LB}</math>
Thus,


<cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath>
<cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath>


Now let apply angle bisector thm.
Now lets apply the angle bisector theorem.


<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>
<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>

Revision as of 01:51, 21 March 2009

Problem

Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$, and $\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$, and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$. If $AM = 180$, find $LP$.

Solution

Sorry, I failed to get the diagram up here, someone help me.

Since $K$ is the midpoint of $\overline{PM}, \overline{AC}$.

Thus, $AK=CK,PK=MK$ and the opposite angles are congruent.

Therefore, $\bigtriangleup{AMK}$ is congruent to $\bigtriangleup{CPK}$ because of SAS

$\angle{KMA}$ is congruent to $\angle{KPA}$ because of CPCTC

That shows $\overline{AM}$ is parallel to $\overline{CP}$ (also $CL$)

That makes $\bigtriangleup{AMB}$ congruent to $\bigtriangleup{LPB}$

Thus,

\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]

Now lets apply the angle bisector theorem.

\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]

\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]

\[\frac {180}{LP}=\frac {5}{2}\]

\[LP=\boxed {072}\]

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_5&oldid=30913"