Difference between revisions of "2011 AIME I Problems/Problem 4"

(Solution: Nitpicking. Also previous post was done as it is clear that the assertion of OQ being a midline and MQC, NOC being isosceles was not addressed as some have asked why the facts were true)
Line 23: Line 23:
== See also ==
== See also ==
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}
{{AIME box|year=2011|n=I|num-b=3|num-a=5}}
{{MAA Notice}}

Revision as of 20:24, 4 July 2013

Problem 4

In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$.


Extend ${MN}$ such that it intersects lines ${AC}$ and ${BC}$ at points $O$ and $Q$, respectively.

Lemma 1: $O, Q$ are midpoints of $AC$ and $BC$

Proof: Consider the reflection of the vertex $C$ over the line $BM$, and let this point be $C_1$. Since $\angle{BMC} = 90^{\circ}$, we have that $C_1$ is the image of $C$ after reflection over $M$, and from the definition of reflection $\angle{MBC} = \angle{MBC_1}$. Then it is easily seen that since $BM$ is an angle bisector, that $\angle{MBC_1} = \angle{MBA}$, so $C_1$ lies on $AB$. Similarly, if we define $C_2$ to be the reflection of $C$ over $N$, then we find that $C_2$ lies on $AB$. Then we can now see that $\triangle{CMN} \sim \triangle{CC_1C_2}$, with a homothety of ratio $2$ taking the first triangle to the second. Then this same homothety takes everything on the line $MN$ to everything on the line $AB$. So since $O, Q$ lie on $MN$, this homothety also takes $O, Q$ to $A, B$ so they are midpoints, as desired. $\Box$

Lemma 2: $\triangle{MQC}, \triangle{NOC}$ are isosceles triangles

Proof: To show that $\triangle{MQC}$ is isosceles, note that $\triangle{MQC} \sim \triangle{C_1BC}$, with similarity ratio of $\frac{1}{2}$. So it suffices to show that triangle $\triangle{C_1BC}$ is isosceles. But this follows quickly from Lemma 1, since $BM$ is both an altitude and an angle bisector of $\angle{C_1BC}$. $\triangle{NOC}$ is isosceles by the same reasoning. $\Box$

Since ${OQ}$ is a midline, it then follows that ${OC} = 58.5$ and ${QC} = 60$. Since $\triangle MQC$ and $\triangle NOC$ are both isosceles, we have that $ON = OC = 58.5$ and $MQ = QC = 60$. Since $OQ$ is a midline, $OQ = 62.5$. We want to find $MN$, which is just $ON + MQ - OQ$.

Substituting the values of $ON, MQ, OQ$, we have that the answer is $58.5 + 60 - 62.5 = \boxed {56}$.

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS