IMO 1960

Key idea: the divisibility rule for $11$ splits the problem into two cases. In each case, a parity check and a discriminant leave only a few digits to try.

Let the digits be $a,b,c$ with $a\ge1$. The number $100a+10b+c$ is divisible by $11$ exactly when $a-b+c$ is a multiple of $11$. Since $-9\le a-b+c\le18$, that means $a-b+c=0$ or $11$.

Case $b=a+c$. The number is $100a+10(a+c)+c=11(10a+c)$, so the quotient is $10a+c$. The condition is $$10a+c=a^2+(a+c)^2+c^2=2a^2+2ac+2c^2 .$$ The right side is even, so $c$ is even. As a quadratic in $a$, $2a^2+(2c-10)a+(2c^2-c)=0$, with discriminant $(2c-10)^2-8(2c^2-c)=100-32c-12c^2$. This is negative for $c\ge2$, so $c=0$. Then $2a^2-10a=0$ gives $a=5$ and $b=5$. That is the number $550$. Check: $550=11\cdot50$ and $25+25+0=50$.

Case $b=a+c-11$. The number is $110a+11c-110=11(10a+c-10)$, so the condition is $$10a+c-10=a^2+(a+c-11)^2+c^2 .$$ Since $t^2\equiv t\pmod 2$, the right side has the parity of $a+(a+c-11)+c$, which is odd. So $c$ is odd. As a quadratic in $a$: $$2a^2+(2c-32)a+\bigl[(c-11)^2+c^2-c+10\bigr]=0 .$$ A quarter of the discriminant is $(c-16)^2-2\bigl[(c-11)^2+c^2-c+10\bigr]=-3c^2+14c-6$. It is non-negative only for $0.5<c<4.2$, so $c=1$ or $c=3$.

• $c=1$: $a^2-15a+55=0$, whose discriminant $5$ is not a square. No digit.
• $c=3$: $a^2-13a+40=0$, so $a=5$ or $a=8$. Then $b=a-8$ must be a digit, so $a=8$ and $b=0$. That is $803=11\cdot73$, and $64+0+9=73$.

The numbers are $550$ and $803$.

Key idea: with $s=\sqrt{1+2x}$, the fraction simplifies to a perfect square.

The root needs $x\ge-\tfrac12$, and the denominator vanishes at $x=0$, so the domain is $x\ge-\tfrac12$, $x\ne0$. Put $s=\sqrt{1+2x}$, so $s\ge0$, $s\ne1$ and $2x=s^2-1$. Then $$\frac{4x^2}{(1-s)^2}=\frac{(s^2-1)^2}{(s-1)^2}=(s+1)^2 .$$ The right side of the inequality is $2x+9=s^2+8$. So the inequality becomes $$s^2+2s+1<s^2+8\iff s<\tfrac72\iff 1+2x<\tfrac{49}{4}\iff x<\tfrac{45}{8}.$$ Together with the domain, the answer is $-\tfrac12\le x<\tfrac{45}8$, $x\ne0$.

Key idea: the middle piece is centered at the midpoint of the hypotenuse, and the right-angle vertex is at distance $a/2$ from that midpoint. Coordinates then give the tangent at once.

Take $n\ge3$ odd (for $n=1$ the “piece” is the whole hypotenuse and the angle is $90^\circ$). Put the hypotenuse on the $x$-axis with its midpoint $O$ at the origin. The middle piece has the same midpoint, so its ends are $(\pm\delta,0)$ with $\delta=\frac{a}{2n}$. The right-angle vertex is $V=(u,h)$, where $h$ is the altitude and $u^2+h^2=VO^2=\frac{a^2}{4}$, because $V$ lies on the circle with the hypotenuse as diameter.

The vectors from $V$ to the two ends are $(-\delta-u,\,-h)$ and $(\delta-u,\,-h)$. Their cross product has size $$|(-\delta-u)(-h)-(-h)(\delta-u)|=2h\delta,$$ and their dot product is $$(-\delta-u)(\delta-u)+h^2=u^2+h^2-\delta^2=\frac{a^2}{4}-\frac{a^2}{4n^2}>0 .$$ The tangent of the angle between two vectors is the size of the cross product over the dot product (the angle is acute because the dot product is positive). So $$\tan\alpha=\frac{2h\delta}{\frac{a^2}{4}\bigl(1-\frac1{n^2}\bigr)}=\frac{ha/n}{a^2(n^2-1)/(4n^2)}=\frac{4nh}{(n^2-1)\,a}.$$

Key idea: halve the second altitude. The midpoint $M$ of $BC$ is at distance $h_b/2$ from line $AC$, so line $AC$ is a tangent from $A$ to the circle about $M$ with radius $h_b/2$.

Step 1. Let $H$ be the foot of the altitude from $A$ and $M$ the midpoint of $BC$. The triangle $AHM$ has a right angle at $H$, leg $AH=h_a$ and hypotenuse $AM=m_a$. Construct it (if $m_a=h_a$, then $H=M$). Line $HM$ is the line $BC$.

Step 2. The distance from $B$ to line $AC$ is $h_b$. Since $C$ lies on line $AC$ and $M$ is the midpoint of $BC$, the distance from $M$ to line $AC$ is half of it, $h_b/2$. So line $AC$ is a line through $A$ tangent to the circle $k$ with center $M$ and radius $h_b/2$. Draw the tangents from $A$ to $k$.

Step 3. Each tangent meets line $HM$ at a point $C$. Reflect $C$ in $M$ to get $B$. The triangle $ABC$ has altitude $h_a$ from $A$ (the distance from $A$ to line $HM$), median $m_a$, and distance $h_b$ from $B$ to line $AC$ (twice the distance from $M$). So it is a solution, and every solution arises this way.

Discussion.

• Step 1 needs $h_a\le m_a$.
• Step 2 needs $A$ outside or on $k$: $m_a\ge h_b/2$. If $m_a>h_b/2$ there are two tangents; if $m_a=h_b/2$ there is one.
• Step 3 fails for a tangent parallel to $BC$. Such a tangent is at distance $h_a$ from $BC$, so from $M$, and this happens only when $h_b=2h_a$.

So there are at most two triangles. When $h_a=m_a$, the two tangents are mirror images in line $AM$, and the two triangles are congruent.

Key idea: the point we track depends linearly on the two parameters, so its locus is the image of a square under an affine map: a filled parallelogram, which we identify from its corners.

Take the cube $[0,1]^3$ with the top face at height $1$, so that $A=(0,0,1)$, $B=(1,0,1)$, $C=(1,1,1)$, $D=(0,1,1)$, and each primed vertex lies directly below its partner at height $0$. The top diagonal $AC$ is $X=(s,s,1)$, and the bottom diagonal $B'D'$ is $Y=(1-t,t,0)$, with $0\le s,t\le1$.

(a) The midpoint is $$\tfrac12(X+Y)=\Bigl(\tfrac{1+s-t}{2},\ \tfrac{s+t}{2},\ \tfrac12\Bigr).$$ This is an affine function of $(s,t)$, so as $(s,t)$ fills the square $[0,1]^2$, the midpoint fills the parallelogram spanned by the images of the corners. Those images are $(s,t)=(0,0)\mapsto(\tfrac12,0,\tfrac12)$, $(1,0)\mapsto(1,\tfrac12,\tfrac12)$, $(1,1)\mapsto(\tfrac12,1,\tfrac12)$ and $(0,1)\mapsto(0,\tfrac12,\tfrac12)$. These are the centers of the four side faces. The locus is the filled square with these four corners, lying in the horizontal plane halfway up the cube.

(b) The point $Z$ on $XY$ with $ZY=2XZ$ is $Z=\tfrac13(2X+Y)$: $$Z=\Bigl(\tfrac{2s+1-t}{3},\ \tfrac{2s+t}{3},\ \tfrac23\Bigr).$$ The corners map to $(\tfrac13,0)$, $(1,\tfrac23)$, $(\tfrac23,1)$ and $(0,\tfrac13)$, all at height $\tfrac23$. The edge vectors are $(\tfrac23,\tfrac23)$ and $(-\tfrac13,\tfrac13)$. They are perpendicular, with lengths $\tfrac{2\sqrt2}{3}$ and $\tfrac{\sqrt2}{3}$. So the locus is a filled rectangle in the horizontal plane one third of an edge below the face $ABCD$. Its long sides are parallel to $AC$, and each of its four corners lies on a side face of the cube.

Key idea: express both volumes through the ball’s radius $r$ and the half base angle. The ratio is then $1/(3t^2(1-t^2))$, and $t^2(1-t^2)\le\frac14$.

Cut through the axis. The cone becomes an isosceles triangle with base radius $R$, height $H$ and base angle $\beta$. The ball becomes its incircle, of radius $r$. The incenter lies on the bisector of the base angle, so $r=R\tan\frac\beta2$, and $H=R\tan\beta$. With $t=\tan\frac\beta2\in(0,1)$, we get $R=r/t$ and $\tan\beta=\frac{2t}{1-t^2}$. So $$V_1=\frac\pi3R^2H=\frac\pi3\cdot\frac{r^3}{t^3}\cdot\frac{2t}{1-t^2}=\frac{2\pi r^3}{3t^2(1-t^2)} .$$ The cylinder around the ball has radius $r$ and height $2r$, so $V_2=2\pi r^3$. Hence $$\frac{V_1}{V_2}=\frac1{3\,t^2(1-t^2)} .$$ By AM–GM, $t^2(1-t^2)\le\bigl(\frac{t^2+1-t^2}{2}\bigr)^2=\frac14$, with equality only for $t^2=\frac12$. So $V_1/V_2\ge\frac43$.

(a) The ratio is never $1$, so $V_1\ne V_2$.

(b) The smallest possible $k$ is $\frac43$, reached when $\tan^2\frac\beta2=\frac12$. Then $\cos\beta=\frac{1-t^2}{1+t^2}=\frac13$. The apex angle $\varphi$ (the angle at the vertex subtending a diameter of the base) is $180^\circ-2\beta$, so $\sin\frac\varphi2=\cos\beta=\frac13$, which makes $\varphi\approx38.94^\circ$ (also $\cos\varphi=\frac79$).

Construction of $\varphi$: draw a right triangle with hypotenuse $3$ and one leg $1$. The angle opposite the leg is $\frac\varphi2$; double it.

Key idea: one dot product gives a quadratic for the height of $P$.

Put the base $a$ on the $x$-axis with ends $(\pm\frac a2,0)$ and the base $c$ on the line $y=h$ with ends $(\pm\frac c2,h)$. The axis is the $y$-axis. Let $P=(0,y)$. The leg from $(\frac a2,0)$ to $(\frac c2,h)$ subtends a right angle at $P$ exactly when the vectors from $P$ to its ends are perpendicular: $$\Bigl(\frac a2,-y\Bigr)\cdot\Bigl(\frac c2,h-y\Bigr)=\frac{ac}4-y(h-y)=0 .$$ By symmetry, the other leg gives the same equation. So $$y^2-hy+\frac{ac}{4}=0,\qquad y=\frac{h\pm\sqrt{h^2-ac}}{2}.$$

(a) The points are $(0,y)$ for these roots. (b) The distance from the base $a$ is $y$ and from the base $c$ is $h-y$. The two roots are swapped by $y\leftrightarrow h-y$, so the two points are placed symmetrically between the bases.

(c)

• $h^2>ac$: two points, both strictly between the bases (both roots lie in $(0,h)$, since their sum is $h$ and their product is positive).
• $h^2=ac$: one point, halfway between the bases. The circle with a leg as diameter then just touches the axis.
• $h^2<ac$: no point.

Geometrically, $P$ must lie on the circle with a leg as diameter. That circle has its center at distance $\frac{a+c}{4}$ from the axis and radius half the leg. The condition $h^2\ge ac$ is the same as $(a+c)^2\le(a-c)^2+4h^2$, which says this circle reaches the axis.