AIME 2009 I · 第 5 题

Solution 1

Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$

Thus,

$$\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1$$ Now let's apply the angle bisector theorem.

$$\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}$$ $$\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}$$ $$\frac {180}{LP}=\frac {5}{2}$$ $$LP=\boxed {072}$$

Note: Another way to realize that $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$ is through angle chasing and congruent triangles. Angle $AKM$ is congruent to angle $CKP$. Segments $MK$ and $PK$ are congruent as stated in the problem. Segments $AK$ and $CK$ are congruent as stated in the problem. Thus, by Side-Angle-Side congruency $\bigtriangleup{AMK}$ is similar to $\bigtriangleup{CPK}$. Then angle $AMK = AMB$ is congruent to angle $CPK = BPL$. Notice triangles $\bigtriangleup{AMB}$ and $\bigtriangleup{LPB}$ both share angle $MBA$. Thus the two triangles are similar because of AA similarity with angles [$AMB$ and $LPB$] and [$MBA = PBL$]. *Solution in Beauty of Math video written out

Solution 2 (Mass Points)

Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem:

$$\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL$$ So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$. Since $K$ is the midpoint of $A$ and $C$, the weight of $A$ is equal to the weight of $C$, which equals $2$. Also, since the weight of $L$ is $5$ and $C$ is $2$, we can weight $P$ as $7$.

By the definition of mass points,

$$\frac{LP}{CP}=\frac{2}{5}\implies LP=\frac{2}{5}CP$$ By vertical angles, angle $MKA =$ angle $PKC$. Also, it is given that $AK=CK$ and $PK=MK$.

By the SAS congruence, $\triangle MKA$ = $\triangle PKC$. So, $MA$ = $CP$ = $180$. Since $LP=\frac{2}{5}CP$, $LP = \frac{2}{5}(180) = \boxed{072}$

Solution 3 (Law of Cosines Bash)

Using the diagram from solution $1$, we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$.

Applying the angle bisector theorem to $\triangle CKB$, we get that $\frac{KP}{PB} = \frac{225}{300} = \frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$.

Now, apply law of cosines on $\triangle CKP$ and $\triangle CPB.$

If we let $\angle KCP = \angle PCB = \alpha$, then the law of cosines gives the following system of equations:

$$9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha$$ $$16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos \alpha.$$ Bashing those out, we get that $x = 15 \sqrt{13}$ and $\cos \alpha = \frac{7}{10}.$

Since $\cos \alpha = \frac{7}{10}$, we can use the double angle formula to calculate that $\cos \left(2 \alpha \right) = -\frac{1}{50}.$

Now, apply Law of Cosines on $\triangle ABC$ to find $AB$.

We get:

$$AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).$$ Bashing gives $AB = 30 \sqrt{331}.$

From the angle bisector theorem on $\triangle ABC$, we know that $\frac{AL}{BL} = \frac{450}{300} = \frac{3}{2}.$ So, $AL = 18 \sqrt{331}$ and $BL = 12 \sqrt{331}.$

Now, we apply Law of Cosines on $\triangle ALC$ and $\triangle BLC$ in order to solve for the length of $LC$.

We get the following system:

$$(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \cdot LC \cdot \frac{7}{10}$$ $$(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}$$ The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$.

The only value that satisfies both equations is $LC = 252$, and since $LP = LC - PC$, we have

Video Solution

https://youtu.be/2Xzjh6ae0MU

~IceMatrix

Video Solution

https://youtu.be/kALrIDMR0dg

~Shreyas S

Solution 4(Area Ratios)

Note that we are given that $\overline{MK} = \overline{KP}$, that $\overline{AK} = \overline{CK}.$ Note then that $\angle MKA = \angle CKB$ by vertical angles. From this, we have $\triangle MKA \cong PKC.$ This means that $\overline{CP}$ is 180. Applying angle bisector theorem on $\triangle ACB$ gives $\frac{\overline{AL}}{\overline{LB}} = \frac{450}{300} = \frac{3}{2}.$ Applying it on $\triangle KCB$ yields

$$\frac{\overline{KP}}{\overline{PB}} = \frac{225}{300} = \frac{3}{4}$$ Now we can proceed with area ratios. Suppose the area of $\triangle ACB = A.$ This means that

$$[\triangle AKL] = \left(\frac{225}{225+225}\right)\left(\frac{3}{5}\right)A = \frac{3}{10}A$$ Continuing on $\triangle LPB$ we have

$$[\triangle LPB] = \left(\frac{2}{2+3}\right)\left(\frac{1}{2}\right)\left(\frac{4}{4+3}\right) = \frac{4}{35}A$$ Since $\overline{AK}=\overline{KC}$ $[\triangle KPL] = [\triangle AKB]-[\triangle AKL] - [\triangle LPB] = \frac{1}{2}A - \frac{3}{10}A - \frac{4}{35}A = \frac{3}{35}A.$ Area ratios on $\triangle KCP$ yield $[\triangle KCP] = \left(\frac{1}{2}\right)\left(\frac{3}{3+4}\right) = \frac{3}{14}.$ Now, suppose $\overline{LP} = x.$ We have that the ratio of areas of $\triangle LKP$ and $\triangle PKC$ is $\frac{x}{180}$ and is also $\frac{\frac{3}{35}}{\frac{3}{14}}$ and equating these gives