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

Revision as of 08:18, 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 $$ (Error compiling LaTeX. Unknown error_msg)\angle BAF + \angle ABC = 90^{\circ} $.

Now notice that$ (Error compiling LaTeX. Unknown error_msg)\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 27: / Line 27: @@
 Denote the midpoint of <math>PQ </math> as <math>R </math>.  It is clear that <math>R </math>'s distance from <math>AB </math> is the average of the distances of <math>P </math> and <math>Q </math> from <math>AB </math>, i.e., half the length of <math>AB</math>, which is a constant.  Therefore the locus in question is a line segment.
+== Solution 2 ==
+[[Image:IMO19595A.png]]
+=== Part a) ===
+Notice that <math>\angle BAF = \arctan (\frac{MF}{AM}) = \frac{MB}{AM} </math> and <math>\angle ABC = </math>arctan (\frac{MC}{MB}) = \frac{AM}{MB} <math>.
-{{alternate solutions}}
+</math>\implies <math> </math>\angle BAF + \angle ABC = 90^{\circ} <math>.
+Now notice that </math>\angle FNB = 90^{\circ} <math>. Considering </math>\angle BAF <math> as </math>\theta <math>, this gives </math>\angle MBN = 90^{\circ} - \theta <math> and thus </math>\angle MFN = 90^{\circ} + \theta <math>. But notice that </math>\angle MFA = 90^{\circ} - \theta <math>, which means that </math>\angle AFN = 180^{\circ} <math>. Therefore points </math>A, F, N <math> are collinear. Now </math>\angle BNF = 90^{\circ} and \angle ANC = 90^{\circ} <math>. Therefore, </math>\angle BNC = 180^{\circ} <math> and thus points </math>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. <math>\angle ANM = \angle ACM = 45^{\circ} </math>. Now <math>\angle ANB = 90^{\circ}</math> which means N lies on the circle with AB as diameter.
+<math>\implies </math> <math>\angle ANX = \angle ABX = 45^{\circ} = \angle ANM </math>. Therefore since M and X are on the same side of <math>N </math>, <math>NM </math> passes through <math>X </math> wherever we choose <math>M </math> on <math>AB </math>.
+=== Part c) ===
+Let the midpoint of <math>PQ </math> be <math>R </math>. Let <math>G </math> be the midpoint of <math>AM </math>. Let <math>F </math> be the midpoint of <math>MB </math>. Let <math>I </math> be the foot of the perpendicular from <math>R </math> onto <math>AB </math>. Therefore by Midpoint Theorem, <math>RI = \frac{PG + HQ}{2} = \frac{AG + HB}{2} = \frac{AB}{4} </math>. Therefore the distance <math>RI </math> is a constant and thus the locus is a straight line parallel to <math>AB </math> at a distance (to <math>AB </math>, of course) of <math>\frac{AB}{4} </math>.
+%alternate solutions%
 == See Also ==

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