IMO 1966

Key idea: name the regions of the Venn diagram. The conditions collapse to one equation in whole numbers, $9c+4m=26$.

Let $a,b,c$ be the numbers who solved only A, only B, only C. Let $m$ be the number who solved B and C but not A, and let $w$ be the number who solved A together with something else.

• Total: $a+b+c+m+w=25$.
• Among those who missed A, solvers of B ($b+m$) are twice the solvers of C ($c+m$): $b+m=2c+2m$, so $b=2c+m$.
• $a=w+1$.
• Half of the single-problem solvers missed A: $b+c=a$.

Then $a=b+c=3c+m$ and $w=a-1$, so the total is $a+a+(a-1)+m=25$, that is $3a+m=26$. Substituting $a=3c+m$ gives $$9c+4m=26 .$$ In non-negative integers: $c=0$ gives $4m=26$ and $c=1$ gives $4m=17$, both impossible; $c=2$ gives $m=2$; and $c\ge3$ is too big. So $c=2$, $m=2$, $a=8$, $w=7$, and $$b=2c+m=6 .$$ Check: $8+6+2+2+7=25$; among those who missed A, $6+2=8$ solved B and $2+2=4$ solved C; and the single-problem solvers are $16$, of whom $8$ missed A.

Key idea: with $\sigma=\frac{\alpha+\beta}2$, the condition factors as $\sin\frac{\beta-\alpha}2\cdot\bigl(\frac{a}{\cos\alpha}-\frac b{\cos\beta}\bigr)=0$. Both factors vanish only when $\alpha=\beta$.

The condition only makes sense when $\alpha,\beta\ne90^\circ$, so both tangents exist. Let $\sigma=\frac{\alpha+\beta}2=90^\circ-\frac\gamma2$. Then $\tan\frac\gamma2=\cot\sigma$, and the condition reads $$a(1-\tan\alpha\cot\sigma)+b(1-\tan\beta\cot\sigma)=0 .$$ Now $$1-\tan\alpha\cot\sigma=\frac{\cos\alpha\sin\sigma-\sin\alpha\cos\sigma}{\cos\alpha\sin\sigma}=\frac{\sin(\sigma-\alpha)}{\cos\alpha\sin\sigma},$$ with $\sigma-\alpha=\frac{\beta-\alpha}2$, and similarly with $\sigma-\beta=-\frac{\beta-\alpha}2$. The condition becomes $$\frac{\sin\frac{\beta-\alpha}2}{\sin\sigma}\Bigl(\frac a{\cos\alpha}-\frac b{\cos\beta}\Bigr)=0 .$$

• Either $\sin\frac{\beta-\alpha}2=0$, which means $\alpha=\beta$,
• or $\frac a{\cos\alpha}=\frac b{\cos\beta}$. By the law of sines $a:b=\sin\alpha:\sin\beta$, so this says $\tan\alpha=\tan\beta$. Two angles of a triangle with equal tangents are equal, since $\tan$ is one-to-one on $(0^\circ,90^\circ)\cup(90^\circ,180^\circ)$ and takes positive values on the first part and negative values on the second.

In both cases $\alpha=\beta$, so the triangle is isosceles with $a=b$.

Key idea: project each distance onto the direction from the center to that vertex. Those four directions cancel.

Let $O$ be the center, $R$ the circumradius, and $u_i=\frac{V_i-O}{R}$ the unit vectors toward the vertices. By symmetry their sum is invariant under every symmetry of the tetrahedron, so it is $0$: $\sum u_i=0$.

For any point $P$, by Cauchy–Schwarz, $$|PV_i|\ge u_i\cdot(V_i-P)=u_i\cdot(V_i-O)-u_i\cdot(P-O)=R-u_i\cdot(P-O).$$ Adding over the four vertices: $$\sum|PV_i|\ge4R-\Bigl(\sum u_i\Bigr)\cdot(P-O)=4R=\sum|OV_i| .$$ Strictness. Equality needs $V_i-P$ to point along $u_i$ for every $i$. That means $P-O=(R-\lambda_i)u_i$ for numbers $\lambda_i\ge0$. Since no two of the $u_i$ are parallel, $P-O$ is parallel to two non-parallel vectors, so $P=O$. For every other point the sum is strictly larger.

Key idea: each term is a difference of two cotangents, so the sum telescopes.

The identity. For $\sin2\theta\ne0$, $$\cot\theta-\cot2\theta=\frac{\cos\theta}{\sin\theta}-\frac{\cos2\theta}{2\sin\theta\cos\theta}=\frac{2\cos^2\theta-\cos2\theta}{\sin2\theta}=\frac1{\sin2\theta},$$ because $2\cos^2\theta-\cos2\theta=1$.

The sum. Apply this with $\theta=2^{k-1}x$ for $k=1,\dots,n$: $$\sum_{k=1}^n\frac1{\sin2^kx}=\sum_{k=1}^n\bigl(\cot2^{k-1}x-\cot2^kx\bigr)=\cot x-\cot2^nx .$$ The excluded values of $x$ are exactly those where some $\sin2^tx$ ($0\le t\le n$) vanishes, so every term is defined.

Key idea: with the $a_i$ in decreasing order, subtracting neighbouring equations removes the absolute values.

How we read the system. The official English text has a misprint in the second equation: its last term repeats the $x_3$ term instead of bringing in $x_4$. We read that term as $|a_2-a_4|\,x_4$, so that all four equations follow the same pattern: equation $i$ is $\sum_{j\ne i}|a_i-a_j|\,x_j=1$. Everything below uses this reading.

Relabel so that $a_1>a_2>a_3>a_4$. Subtract equation 2 from equation 1. For $j=1$ the coefficient is $0-(a_1-a_2)$. For $j\ge2$ it is $(a_1-a_j)-(a_2-a_j)=a_1-a_2$. Dividing by $a_1-a_2>0$: $$-x_1+x_2+x_3+x_4=0 .$$ In the same way, equations 2 minus 3, and 3 minus 4, give $$-x_1-x_2+x_3+x_4=0,\qquad -x_1-x_2-x_3+x_4=0 .$$ The difference of the last two gives $x_3=0$. The difference of the first two gives $x_2=0$. Then $x_1=x_4$. Equation 1 now reads $(a_1-a_4)x_4=1$. So $$x_1=x_4=\frac1{a_1-a_4},\qquad x_2=x_3=0 .$$ Check equation 2: $(a_1-a_2)x_1+(a_2-a_4)x_4=\frac{a_1-a_4}{a_1-a_4}=1$, and equation 3 the same way. The solution is unique.

In the original labels: the unknowns belonging to the largest and the smallest $a$ both equal $1/(a_{\max}-a_{\min})$, and the other two are $0$.

Key idea: the product of the three corner areas is a product of three factors $t(1-t)$, each at most $\frac14$.

Let $S$ be the area of $ABC$, and put $AM=m\cdot AB$, $BK=k\cdot BC$, $CL=l\cdot CA$ with $m,k,l\in(0,1)$. A triangle that shares an angle with $ABC$ has area proportional to the product of the two sides along that angle. So $$\frac{[AML]}S=m(1-l),\qquad\frac{[BKM]}S=k(1-m),\qquad\frac{[CLK]}S=l(1-k).$$ Multiplying, $$\frac{[AML][BKM][CLK]}{S^3}=m(1-m)\cdot k(1-k)\cdot l(1-l)\le\Bigl(\frac14\Bigr)^3,$$ since $t(1-t)\le\frac14$ for every $t$. If all three areas were larger than $\frac S4$, the product would be larger than $(\frac S4)^3$. So at least one area is at most $\frac S4$. (Equality throughout happens when all three points are midpoints.)