# International Mathematical Olympiad '59, Second day

The goal of this set of notes is to improve our problem solving and proof writing skills. You are encouraged to submit a solution to any of these problems, and join in the discussion in #imo-discussion on Saturday at 9:00 pm IST ,8 30 PDT. For more details, see IMO Problems Discussion Group.

Here are the problems from the second day of the 1959 International mathematical Olympiad. They range from easy to harder Geometry and Combinatorics. Try your hand at them. Don't be discouraged if you cant completely solve them. Do post your inspirations and ideas towards the problems. The discussion for theses questions will be held soon. Happy Problem Solving!

Q4. Construct a right-angled triangle whose hypotenuse $c$ is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle. (HUN)

Q5. A segment $AB$ is given and on it a point $M$. On the same side of $AB$ squares $AMCD$ and $BMFE$ are constructed. The circumcircles of the two squares, whose centers are $P$ and $Q$, intersect in $M$ and another point $N$.

(a) Prove that lines $FA$ and $BC$ intersect at $N$.

(b) Prove that all such constructed lines $MN$ pass through the same point $S$, regardless of the selection of $M$.

(c) Find the locus of the midpoints of all segments $PQ$, as $M$ varies along the segment $AB$.(ROM)

Q6. Let $\alpha$ and $\beta$ be two planes intersecting at a line $p$. In $\alpha$ a point $A$ is given and in $\beta$ a point $C$ is given, neither of which lies on $p$. Construct $B$ in $\alpha$ and $D$ in $\beta$ such that $ABCD$ is an equilateral trapezoid, $AB || CD$, in which a circle can be inscribed. (CZS)

###### This is part of the set International Mathematical Olympiad Note by Sualeh Asif
5 years, 10 months ago

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$Q1.$ 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 $\sqrt{ab}$.

$\:$

$Lemma:$ 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 $D$. Alt text

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.

$\:$

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

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

$\:$

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)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.

- 5 years, 10 months ago

Good work in finding the angles of such a triangle! Now ponder on this: In Euclidean Geometry when you are asked to construct a figure you are expected to use a straight edge and a compass only. @Shamay Samuel your job is to find a way to construct this triangle. This solution wont be valid at the IMO. Since it does not give a way to construct this triangle. So find a way to complete this solution!

Secondly, In the third line you say " We know that...." As Sir @Calvin Lin pointed out this is not acceptable at proof-writing. Please write why you know that, and how the others can understand it!

However great start! try and finish it.

- 5 years, 10 months ago

Thanks, @Sualeh Asif for your advice! I have made changes to my solution according to it. Please check it to see if it can be improved upon.

- 5 years, 10 months ago

@Shamay Samuel can you define how to make an angle of $60^{\circ}$. The solution is still incomplete.That would be another problem. Great proving the firs fact.

- 5 years, 10 months ago

Good clear presentation. Keep it up!

At the IMO, this will be worth 7-.

A slightly easier way to construct a $15 ^ \circ$ angle is to construct $60^\circ$ through an equilateral triangle and then $45^\circ$ through a right isosceles triangle.

Staff - 5 years, 10 months ago

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.

- 5 years, 10 months ago

In a), what is $N'$? Unfortunately, as I am not a mind-reader, your solution doesn't make sense.

In b), where did that first equation come from? You should also prove why "the bisector of an arc is a constant point", and it helps to state what that point is.

In c), you have to specify what is the exact locus, and not just what the shape of the locus is. IE Give the exact endpoints, and prove that they can be achieved, and that no other points can be achieved.

At the IMO, this solution is worth 0+. It is a classic case of where you as the solver knows enough to completely solve the problem, but when writing up the solution you miss out crucial bits of information and explanation which completely destroys the building of your proof.

Staff - 5 years, 10 months ago

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.

- 5 years, 10 months ago

Good work. For c, note that you're just asked to find the locus of the points, and do not need to give a geometric interpretation of it. Defining the locus is the crucial part here.

Staff - 5 years, 10 months ago

@salmaan shahid This is not a solution, Please write a complete solution.

- 5 years, 10 months ago