Difference between revisions of "1959 IMO Problems/Problem 5"

Revision as of 07:17, 23 May 2024

Problem

An arbitrary point $M$ is selected in the interior of the segment $AB$ . The squares $AMCD$ and $MBEF$ are constructed on the same side of $AB$ , with the segments $AM$ and $MB$ as their respective bases. The circles about these squares, with respective centers $P$ and $Q$ , intersect at $M$ and also at another point $N$ . Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$ .

(a) Prove that the points $N$ and $N'$ coincide.

(b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$ .

(c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$ .

Solution

Part A

Since the triangles $AFM, CBM$ are congruent, the angles $AFM, CBM$ are congruent; hence $AN'B$ is a right angle. Therefore $N'$ must lie on the circumcircles of both quadrilaterals; hence it is the same point as $N$ .

Part B

We observe that $\frac{AM}{MB} = \frac{CM}{MB} = \frac{AN}{NB}$ since the triangles $ABN, BCM$ are similar. Then $NM$ bisects $ANB$ .

We now consider the circle with diameter $AB$ . Since $ANB$ is a right angle, $N$ lies on the circle, and since $MN$ bisects $ANB$ , the arcs it intercepts are congruent, i.e., it passes through the bisector of arc $AB$ (going counterclockwise), which is a constant point.

Part C

Denote the midpoint of $PQ$ as $R$ . It is clear that $R$ 's distance from $AB$ is the average of the distances of $P$ and $Q$ from $AB$ , i.e., half the length of $AB$ , which is a constant. Therefore the locus in question is a line segment.

Solution 2

Part a)

Notice that $\angle BAF =$ arctan \frac{MF}{AM} = \frac{MB}{AM} $and$ \angle ABC = $arctan \frac{MC}{MB} = \frac{AM}{MB}$ .

$\implies$ $\angle BAF + \angle ABC = 90^{\circ}$ .

Now notice that $\angle FNB = 90^{\circ}$ . Considering $\angle BAF$ as $\theta$ , this gives $\angle MBN = 90^{\circ} - \theta$ and thus $\angle MFN = 90^{\circ} + \theta$ . But notice that $\angle MFA = 90^{\circ} - \theta$ , which means that $\angle AFN = 180^{\circ}$ . Therefore points $A, F, N$ are collinear. Now $\angle BNF = 90^{\circ} and \angle ANC = 90^{\circ}$ . Therefore, $\angle BNC = 180^{\circ}$ and thus points $B, N, C$ are collinear. Therefore, AF and BC intersect at N.

Part b)

Construct the bisector of arc AB above AB. Call it X. $\angle ANM = \angle ACM = 45^{\circ}$ . Now $\angle ANB = 90^{\circ}$ which means N lies on the circle with AB as diameter.

$\implies$ $\angle ANX = \angle ABX = 45^{\circ} = \angle ANM$ . Therefore since M and X are on the same side of $N$ , $NM$ passes through $X$ wherever we choose $M$ on $AB$ .

Part c)

Let the midpoint of $PQ$ be $R$ . Let $G$ be the midpoint of $AM$ . Let $F$ be the midpoint of $MB$ . Let $I$ be the foot of the perpendicular from $R$ onto $AB$ . Therefore by Midpoint Theorem, $RI = \frac{PG + HQ}{2} = \frac{AG + HB}{2} = \frac{AB}{4}$ . Therefore the distance $RI$ is a constant and thus the locus is a straight line parallel to $AB$ at a distance (to $AB$ , of course) of $\frac{AB}{4}$ .

%alternate solutions%

@@ Line 30: / Line 30: @@
 [[Image:IMO19595A.png]]
 === Part a) ===
-Notice that <math>\angle BAF = </math>\taninv \frac{MF}{AM} = \frac{MB}{AM} <math> and </math>\angle ABC = <math>taninv \frac{MC}{MB} = \frac{AM}{MB} </math>.
+Notice that <math>\angle BAF = </math>arctan \frac{MF}{AM} = \frac{MB}{AM} <math> and </math>\angle ABC = <math>arctan \frac{MC}{MB} = \frac{AM}{MB} </math>.
 <math>\implies </math> <math>\angle BAF + \angle ABC = 90^{\circ} </math>.

1959 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions

Page

Toolbox

Search