International Mathematical Olympiad '59, Second day

$Q1.$ Let the legs of the right triangle $\Delta ABC$ (right-angled at $C$) be $a$ & $b$.

The median from $C$ is defined to have a length of $\sqrt{ab}$.

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$Lemma:$ The length of the median of a right triangle, from the right angle to the hypotenuse, is half the length of the hypotenuse.

$Proof:$ Suppose we have a right triangle $\Delta ABC$ (right-angled at $C$). Let the mid-point of $AB$ be $D$.

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We construct a line $\parallel$ to $AB$ from $D$. Let it intersect $BC$ at $E$.

Since $DE \parallel AB \Rightarrow \angle DEC = \angle DEB = 90^\circ$

Also, $EB=EC$ (Mid-point Theorem) and $DE$ is common to both $\Delta DEB$ & $\Delta DEC$

$\therefore \Delta DEB \cong \Delta DEC$

$\Rightarrow DA = DB = DC$

Hence, proved.

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We now use the above statement to proceed,

$\therefore \sqrt{ab} = \frac{c}{2} \Rightarrow ab = \frac{c^2}{4}$

$\therefore 2 \cdot \frac{a}{c} \cdot \frac{b}{c} = \frac{1}{2}$

$\therefore 2\sin A \cos A = \frac{1}{2} \Rightarrow \sin 2A = \frac{1}{2}$

Since, $0 \leq 2A \leq \pi$, $\Rightarrow 2A = \frac{\pi}{6}, \frac{5\pi}{6}$

$\therefore A = \frac{\pi}{12}, \frac{5\pi}{12} \Leftrightarrow A = 15^\circ, 75^\circ$

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In order to construct such a triangle, we will need to construct $60^\circ$ angles. To do so, we construct a segment (say, $QR$). We take a radius less than the length of $QR$ and construct an arc, with center $Q$, intersecting $QR$ at $R'$. Using the same radius and with center $R'$, we construct an arc intersecting the first arc at $P$. $\angle PQR$ is the required angle. (We can construct a $120^\circ$ angle by intersecting the first arc again, with center $P$ and the same radius, as before)

We also need to bisect angles. To do so, we construct an arbitrary angle (say, $\angle PQR$). We take a radius less than the length of $QR$ (or $QP$, if $QP<QR$)and construct an arc, with center $Q$, intersecting $QR$ and $QP$ at $R'$ and $P'$, respectively. Using $R'$ and $P'$ as centers, using the same radius, we construct an arc from both of them. These arcs intersect at two points, which we join and extend to $Q$ to construct the angle bisector of $\angle PQR$.

Using the above two methods, we can now construct such a right triangle $\Delta ABC$ with these angles.

We construct a segment $AB$ with the given length $c$. On $A$, we construct a $60^\circ$ angle and bisect it twice, towards $AB$ giving us $15^\circ$. On $B$, we construct a $120^\circ$ angle and bisect the angle between the $60^\circ$ and the $120^\circ$ mark twice, towards the $60^\circ$ mark giving us $75^\circ$. We extend the new segments and label their intersection point $C$, which is right-angled. Thus, $\Delta ABC$ is the required triangle.

2) This is my attempt at Question 2

(a)

The triangles $AFM$ and $CBM$ are congruent by the $SAS$ criterion ($AM = CM$, $FM=BM$ and $\angle AMF = \angle BMC$). This implies that $\Delta CBM$ is similar to $\Delta CN'F$ by the $AA$ criterion ($\angle CBM = \angle CFN'$ and $\angle BCM = \angle FCN'$). This means that $\angle CMB = \angle CN'F = \angle CN'A = 90^{\circ}$. Since $\angle FN'B + \angle FEB = 180^{\circ}$ (because they are both $90^{\circ}$, $FN'BE$ is a cyclic quadrilateral. Furthermore, since $\angle CN'A + \angle CMA = 180^{\circ}$ (because they are both $90^{\circ}$, $CN'MA$ is a cyclic quadrilateral. Therefore, $N'$ lies on both circumcircles; hence it is the same point as $N$.

(b)

Note that $\dfrac {AM}{MB} = \dfrac {CM}{MB} = \dfrac {AN}{NB}$. Thus, by the converse of angle bisector theorem, $NM$ is the angle bisector of $\angle ANB$

Consider a circle with diameter $AB$. It follows that point $N$ is on the circumference since $\angle ANB = 90^{\circ}$ (Angle in semicircle). The angle $\angle ANB$ subtends arc $AB$ (This arc has not got point $N$ on it). Since $MN$ bisects $\angle ANB$, $MN$ must also bisect the arc $AB$. Since the bisector of an arc is a constant point, the line $MN$ passes through a constant point.

(c)

The average of the distances of $P$ and $Q$ from $AB$ gives the distance of the midpoint of $PQ$ from $AB$. Since $P$ is a distance of $\dfrac {1}{2} AM$ and $Q$ is distance of $\dfrac {1}{2} MB$, the midpoint is a distance of $\dfrac {1}{4} AB$ from the line $AB$, which is a constant. Therefore, the locus of this midpoint is a line segment.

2nd one is quite easy using coordinates. Fix A as origin. a) First one is simple, just solve some simple equations and we are done. b) For 2nd one use the idea that equation of radical axis of two circles is S1 - S2 . c) For 3rd one writing the coordinates of mid-point we see that it's y coordinate is always fixed. But I have not yet figured out what the locus of x coordinate means geometrically.

I will post my solution later as I have my exam tomorrow.