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 cc 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 ABAB is given and on it a point MM. On the same side of ABAB squares AMCDAMCD and BMFEBMFE are constructed. The circumcircles of the two squares, whose centers are PP and QQ, intersect in MM and another point NN.

(a) Prove that lines FAFA and BCBC intersect at NN.

(b) Prove that all such constructed lines MNMN pass through the same point SS, regardless of the selection of MM.

(c) Find the locus of the midpoints of all segments PQPQ, as MM varies along the segment ABAB.(ROM)

Q6. Let α\alpha and β\beta be two planes intersecting at a line pp. In α\alpha a point AA is given and in β\beta a point CC is given, neither of which lies on pp. Construct BB in α\alpha and DD in β\beta such that ABCDABCD is an equilateral trapezoid, ABCDAB || CD, in which a circle can be inscribed. (CZS)

This is part of the set International Mathematical Olympiad

Note by Sualeh Asif
3 years, 10 months ago

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Q1.Q1. Let the legs of the right triangle ΔABC\Delta ABC (right-angled at CC) be aa & bb.

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

\:

Lemma: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:Proof: Suppose we have a right triangle ΔABC\Delta ABC (right-angled at CC). Let the mid-point of ABAB be DD.

Alt text Alt text

We construct a line \parallel to ABAB from DD. Let it intersect BCBC at EE.

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

Also, EB=ECEB=EC (Mid-point Theorem) and DEDE is common to both ΔDEB\Delta DEB & ΔDEC\Delta DEC

ΔDEBΔDEC\therefore \Delta DEB \cong \Delta DEC

DA=DB=DC\Rightarrow DA = DB = DC

Hence, proved.

\:

We now use the above statement to proceed,

ab=c2ab=c24\therefore \sqrt{ab} = \frac{c}{2} \Rightarrow ab = \frac{c^2}{4}

2acbc=12\therefore 2 \cdot \frac{a}{c} \cdot \frac{b}{c} = \frac{1}{2}

2sinAcosA=12sin2A=12\therefore 2\sin A \cos A = \frac{1}{2} \Rightarrow \sin 2A = \frac{1}{2}

Since, 02Aπ0 \leq 2A \leq \pi, 2A=π6,5π6\Rightarrow 2A = \frac{\pi}{6}, \frac{5\pi}{6}

A=π12,5π12A=15,75\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 6060^\circ angles. To do so, we construct a segment (say, QRQR). We take a radius less than the length of QRQR and construct an arc, with center QQ, intersecting QRQR at RR'. Using the same radius and with center RR', we construct an arc intersecting the first arc at PP. PQR\angle PQR is the required angle. (We can construct a 120120^\circ angle by intersecting the first arc again, with center PP and the same radius, as before)

We also need to bisect angles. To do so, we construct an arbitrary angle (say, PQR\angle PQR). We take a radius less than the length of QRQR (or QPQP, if QP<QRQP<QR)and construct an arc, with center QQ, intersecting QRQR and QPQP at RR' and PP', respectively. Using RR' and PP' 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 QQ to construct the angle bisector of PQR\angle PQR.

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

We construct a segment ABAB with the given length cc. On AA, we construct a 6060^\circ angle and bisect it twice, towards ABAB giving us 1515^\circ. On BB, we construct a 120120^\circ angle and bisect the angle between the 6060^\circ and the 120120^\circ mark twice, towards the 6060^\circ mark giving us 7575^\circ. We extend the new segments and label their intersection point CC, which is right-angled. Thus, ΔABC\Delta ABC is the required triangle.

Shamay Samuel - 3 years, 10 months ago

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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.

Sualeh Asif - 3 years, 10 months ago

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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.

Shamay Samuel - 3 years, 10 months ago

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@Shamay Samuel @Shamay Samuel can you define how to make an angle of 6060^{\circ}. The solution is still incomplete.That would be another problem. Great proving the firs fact.

Sualeh Asif - 3 years, 10 months ago

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Good clear presentation. Keep it up!

At the IMO, this will be worth 7-.

A slightly easier way to construct a 15 15 ^ \circ angle is to construct 6060^\circ through an equilateral triangle and then 45 45^\circ through a right isosceles triangle.

Calvin Lin Staff - 3 years, 10 months ago

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2) This is my attempt at Question 2

(a)

The triangles AFMAFM and CBMCBM are congruent by the SASSAS criterion (AM=CMAM = CM, FM=BMFM=BM and AMF=BMC\angle AMF = \angle BMC). This implies that ΔCBM\Delta CBM is similar to ΔCNF\Delta CN'F by the AAAA criterion (CBM=CFN\angle CBM = \angle CFN' and BCM=FCN\angle BCM = \angle FCN'). This means that CMB=CNF=CNA=90\angle CMB = \angle CN'F = \angle CN'A = 90^{\circ}. Since FNB+FEB=180\angle FN'B + \angle FEB = 180^{\circ} (because they are both 9090^{\circ}, FNBEFN'BE is a cyclic quadrilateral. Furthermore, since CNA+CMA=180\angle CN'A + \angle CMA = 180^{\circ} (because they are both 9090^{\circ}, CNMACN'MA is a cyclic quadrilateral. Therefore, NN' lies on both circumcircles; hence it is the same point as NN.

(b)

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

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

(c)

The average of the distances of PP and QQ from ABAB gives the distance of the midpoint of PQPQ from ABAB. Since PP is a distance of 12AM\dfrac {1}{2} AM and QQ is distance of 12MB\dfrac {1}{2} MB, the midpoint is a distance of 14AB\dfrac {1}{4} AB from the line ABAB, which is a constant. Therefore, the locus of this midpoint is a line segment.

Sharky Kesa - 3 years, 10 months ago

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In a), what is NN'? 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.

Calvin Lin Staff - 3 years, 10 months ago

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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.

salmaan shahid - 3 years, 10 months ago

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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.

Calvin Lin Staff - 3 years, 10 months ago

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

Sualeh Asif - 3 years, 10 months ago

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