Monday, March 29, 2021

1960 IMO Problems And Solutions

 

Problem 01

Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$.

Solutions

Solution 1

Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$. Then\[100a + 10b+c = 11m\]for some $m$. We also have $m=a^2+b^2+c^2$. Substituting this into the first equation and simplification, we get


\[100a+10b+c = 11a^2 +11b^2 +11c^2\]For an integer divisible by $11$, the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by $11$. Thus we get: $b = a + c$ or $b = a + c - 11$.

Case $1$: Let $b=a+c$. We get\[100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2\]\[10a+c = 2a^2+2ac+2c^2\]Since the right side is even, the left side must also be even. Let $c=2q$ for some $q = 0,1,2,3,4$. Then\[10a+2q=2a^2+4aq+8q^2\]\[5a+q=a^2+2aq+4q^2\]Substitute $q=0,1,2,3,4$ into the last equation and then solve for $a$.

When $q=0$, we get $a=5$. Thus $c=0$ and $b=5$. We get that $N=550$ which works.

When $q=1$, we get that $a$ is not an integer. There is no $N$ for this case.

When $q=2$, we get that $a$ is not an integer. There is no $N$ for this case.

When $q=3$, we get that $a$ is not an integer. There is no $N$ for this case.

When $q=4$, we get that $a$ is not an integer. There is no $N$ for this case.

Case $2$: Let $b = a + c - 11$. We get\[100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)\]\[10a+c=2a^2+2c^2+2ac-22a-22c+131\]\[2(a-8)^2+2(c-\frac{23}{4})^2+2ac-\frac{505}{8}=0\]Now we test all $c=0\rightarrow10$. When $c=0,1,2,4,5,6,7,8,9$, we get no integer solution to $a$. Thus, for these values of $c$, there is no valid $N$. However, when $c=3$, we get\[2(a-8)^2+2(3-\frac{23}{4})^2+6a-\frac{505}{8}=0\]\[2(a-8)^2+6a-48 = 0\]We get that $a=8$ is a valid solution. For this case, we get $a=8,b=0,c=3$, so $N=803$, and this is a valid value. Thus, the answers are $\boxed{N=550,803}$.


Solution 2

Define a ten to be all ten positive integers which begin with a fixed tens digit.

We can make a systematic approach to this:


By inspection, $\dfrac{N}{11}$ must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.

For a given ten, the sum of the squares of the digits of $N$ increases faster than $\dfrac{N}{11}$, so we can have at most one number in every ten that works.

We check the first ten:

$11*11=121$

$1^2+2^2+1^2=4$

$12*11=132$

$1^2+3^2+2^2=14$

11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.

We try the second ten:

$21*11=231$

$2^2+3^2+1^2=14$

$22*11=242$

$2^2+4^2+2^2=24$

Therefore, no numbers in the second ten work.

We continue, to find out that 50 and 73 are the only ones that works.

$N=50*11=550$, $N=73*11=803$ so there are two $N$ that works.

Problem 02

For what values of the variable $x$ does the following inequality hold:

\[\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?\]

Solution

Set $x = -\frac{1}{2} + \frac{a^2}{2}$, where $a\ge0$. $\frac{4\left(-\frac{1}{2}+\frac{a^2}{2}\right)^2}{\left(1-\sqrt{1+2\left(-\frac{1}{2}+\frac{a^2}{2}\right)}\right)^2}<2\left(-\frac{1}{2}+\frac{a^2}{2}\right)+9$

After simplifying, we get $(a+1)^2<a^2+8$

So $a^2+2a+1<a^2+8$

Which gives $a<\frac{7}{2}$ and hence $-\frac{1}{2} \le x<\frac{45}{8}$.

But $x=0$ makes the LHS indeterminate.

So, answer: $-\frac{1}{2} \le x<\frac{45}{8}$, except $x=0$.

Problem 03

In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

$\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$

Solution

Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.

[asy] size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); dot(Q); label("A",A,SW); label("B",B,SE); label("C",C,NW); label("P",P,ENE); label("Q",Q,NNE); draw(A--B--C--cycle); draw(A--P); draw(A--Q); [/asy]

Then, $P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)$, and $Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)$.

So, $\text{slope}$$(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}$, and $\text{slope}$$(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}$.

Thus, $\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}$ $= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}$.

Since $[ABC]=\frac{1}{2}bc=\frac{1}{2}ah$, $bc=ah$ and $\tan{\alpha}=\frac{4nh}{(n^2-1)a}$ as desired.

Solution 2

Let $P, Q, R$ be points on side $BC$ such that segment $PR$ contains midpoint $Q$, with $P$ closer to $C$ and (without loss of generality) $AC \le AB$. Then if $AD$ is an altitude, then $D$ is between $P$ and $C$. Combined with the obvious fact that $Q$ is the midpoint of $PR$ (for $n$ is odd), we have\[\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.\]

Problem 04

Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

Solution

Let $M_a$, $M_b$, and $M_c$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let $H_a$, $H_b$, and $H_c$ be the feet of the altitudes from $A$, $B$, and $C$ to their opposite sides, respectively. Since $\triangle ABC\sim\triangle M_bM_aC$, with $M_bM_a=\frac12 AB$, the distance from $M_a$ to side $\overline{AC}$ is $\frac{h_b}{2}$.

Construct $AM_a$ with length $m_a$. Draw a circle centered at $A$ with radius $h_a$. Construct the tangent $l_1$ to this circle through $M_a$. $\overline{BC}$ lies on $l_1$.

Draw a circle centered at $M_a$ with radius $\frac{h_b}{2}$. Construct the tangent $l_2$ to this circle through $A$. $\overline{AC}$ lies on $l_2$. Then $C=l_1\cap l_2$.

Construct the line $l_3$ parallel to $l_2$ so that the distance between $l_2$ and $l_3$ is $h_b$ and $M_a$ lies between these lines. $B$ lies on $l_3$. Then $B=l_1\cap l_3$.

Problem 05

Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$).

a) Find the locus of the midpoints of the segments $XY$, where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$;

b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$.

Solution

Let $A=(0,0,2)$, $B=(2,0,2)$, $C=(2,0,0)$, $D=(0,0,0)$, $A'=(0,2,2)$, $B'=(2,2,2)$, $C'=(2,2,0)$, and $D'=(0,2,0)$. Then there exist real $x$ and $y$ in the closed interval $[0,2]$ such that $X=(x,0,2-x)$ and $Y=(y,2,y)$.

The midpoint of $XY$ has coordinates $((x+y)/2, 1, (2-x+y)/2)$. Let $a$ and $b$ be the $x$- and $z$-coordinates of the midpoint of $XY$, respectively. We then have that $a+b=y+1$ and $a-b=x-1$, so $a+b\in [1,3]$ and $a-b\in [-1,1]$. The region of points that satisfy these inequalities is the closed square with vertices at $(1,1,2)$, $(2,1,1)$, $(1,1,0)$, and $(0,1,1)$. For every point $P$ in this region, there exist unique points $X$ and $Y$ such that $P$ is the midpoint of $XY$.

If $Z\in XY$ and $ZY=2XZ$, then $Z$ has coordinates $((2x+y)/3, 2/3, (4-2x+y)/3)$. Let $a$ and $b$ be the $x$- and $z$- coordinates of $Z$. We then have that $a+b=(4/3)+(2/3)y$ and $a-b=(4x-4)/3$, and $a\in (4/3,8/3)$ and $b\in (-4/3,4/3)$. The region of points that satisfy these inequalities is the closed rectangle with vertices at $(0,2/3,4/3)$, $(2/3,2/3,1)$, $(1,2/3,2/3)$, and $(4/3,2/3,0)$. For every point $Z$ in this region, there exist unique points $X$ and $Y$ such that $Z\in XY$ and $ZY=2XZ$.

Problem 06

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$;

b) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

Solution

Part (a):

Let $R$ denote the radius of the cone, and let $r$ denote the radius of the cylinder and sphere. Let $l$ denote the slant height of the cone, and let $h$ denote the height of the cone.

Consider a plane that contains the axis of the cone. This plane will slice the cone and sphere into a circle $\omega$ inscribed in an isosceles triangle $T$.

The area of $T$ may be computed in two different ways:\[[T] = \frac{1}{2} \times r \times (2l + 2R) = r(l+R)\]\[[T] = \frac{1}{2} \times 2R \times h = Rh\]From this, we deduce that $r = \frac{Rh}{l+R}$.

Now, we calculate our volumes:\[V_1 = \frac{1}{3}\pi R^2 h\]\[V_2 = \pi r^2 \times 2r = 2\pi r^3 = \frac{2R^3 h^3}{(l+R)^3}\]Now, we will compute the quantity $\frac{3(l+R)^3}{\pi R^5 h} (V_1 - V_2)$ and prove that it is always greater than $0$. Let $x = \frac{h}{R}$. Clearly, $x$ can be any positive real number. Define $W_1 = \frac{3(l+R)^3}{\pi R^5 h} V_1$ and $W_2 = \frac{3(l+R)^3}{\pi R^5 h} V_2$. We will calculate $W_1$ and $W_2$ in terms of $x$ and then compute the desired quantity $W_1 - W_2$.

It is easy to see that:\[W_1 = (\sqrt{x^2+1} + 1)^3\]\[W_2 = 6x^2\]

Now, let $u = \sqrt{x^2+1}$. Since $x > 0$, it follows that $u > 1$. We now have:\[W_1 = (u + 1)^3\]\[W_2 = 6(u^2 - 1)\]

Define $f(u) = W_1 - W_2$. It follows that:\[f(u) = (u+1)^3 - 6(u^2 - 1)\]\[f(u) = u^3 - 3u^2 + 3u + 7\]\[f(u) = (u-1)^3 + 8 > 8 > 0\]

We see that $f(u) > 8$ for all allowed values of $u$. Thus, $V_1 - V_2 > \frac{8\pi R^5 h}{3(l+R)^3}$, meaning that $V_1 > V_2$. We have thus proved that $V_1 \ne V_2$, as desired.

Part (b):

From our earlier work in calculating the volumes $V_1$ and $V_2$, we easily see that:\[\frac{V_1}{V_2} = \frac{(u+1)^3}{6(u^2 - 1)}\]Re-expressing and simplifying, we have:\[\frac{6V_1}{V_2} = \frac{(u+1)^2}{u-1} = (u-1) + 4 + \frac{4}{u-1}\]By the AM-GM Inequality, $(u-1) + \frac{4}{u-1} \ge 4$, meaning that $\frac{V_1}{V_2} \ge \frac{4}{3}$. Equality holds if and only if $u-1 = \frac{4}{u-1}$, meaning that $u=3$ and $x = 2\sqrt{2}$.

If we check the case $x = \frac{h}{R} = 2\sqrt{2}$, we may calculate $V_1$ and $V_2$:\[V_1 = \frac{1}{3} \pi R^2 h = \frac{2\sqrt{2}}{3}\pi R^3\]\[V_2 = 2\pi r^3 = 2\pi \left(\frac{R\times 2\sqrt{2} R}{4R}\right)^3 = \frac{\pi}{\sqrt{2}} R^3\]Indeed, we have $\frac{V_1}{V_2} = \frac{4}{3}$, meaning that our minimum of $k = \frac{4}{3}$ can be achieved.

Thus, we have proved that the minimum value of $k$ such that $V_1 = kV_2$ is $\frac{4}{3}$.

Now, let $\theta$ be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following:\[\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}\]From the double-angle formula for tangent,\[\theta = \arctan\left(\frac{4\sqrt{2}}{7}\right)\]This angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths $4\sqrt{2}$ and $7$. This is straightforward, and the angle opposite the leg of length $4\sqrt{2}$ will be the desired angle $\theta$.

It follows that we have successfully constructed the desired angle $\theta$.

Problem 07

An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.

a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;

b) Calculate the distance of $P$ from either base;

c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.

Solution

(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.

(b) Let $x$ be the distance from $P$ to one of the bases; then $h - x$ must be the distance from $P$ to the other base. Similar triangles give $\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}$, so $x^2 - hx + \frac{ac}{4} = 0$ and so $x = \frac{h \pm \sqrt{h^2 - ac}}{2}.$

(c) When $h^2 \ge ac$.

[1]

In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).

Let our point P be on the axis of symmetry at z distance from the origin O.

The coordinates of the points A,B,C,D,E,F and P are given in the figure.

Now,

Slope of the line $PC= (z-0)/(0-c/2) = -2z/c$ Slope of the line $PB= (z-h)/(0-a/2) = -2(z-h)/a$

Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.

i.e

$4z(z-h)=-ac$

or $z^2 - zh + ac/4= 0$

Now, solving for z, we get, $z= [(h + ( h^2 - ac ) ^1/2 ]/2$ and $[(h - ( h^2 - ac ) ^1/2 ]/2$

So, z is the distance of the points from the base CD..

Also the points are possible only when , $h^2 - ac >= 0$.. and doesn't exist for $h^2 -ac <0$

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