IMO 1970 Problem 6
The statement concerns all $\binom{100}{3}$ triangles determined by a set of points in general position.
Proposed by: -
Verified: no
Verdicts: FAIL + UNKNOWN
Solve time: 19m45s
Problem
In a plane there are $100$ points, no three of which are collinear. Consider all possible triangles having these point as vertices. Prove that no more than $70 %$ of these triangles are acute-angled.
Exploration
The statement concerns all $\binom{100}{3}$ triangles determined by a set of points in general position. The target proportion is $70%=\frac 7{10}$. A direct count of acute triangles appears impossible because the geometric configuration is arbitrary.
The first idea is to classify triangles by the angle opposite the longest side. Every nonacute triangle has a unique angle at least $\frac{\pi}{2}$, since the sum of the angles equals $\pi$. Thus each triangle is either acute or has a unique obtuse or right angle. Because no three points are collinear, right triangles may still occur, but each nonacute triangle contributes exactly one distinguished vertex.
A natural way to count acute triangles is to fix a vertex $P$ and count pairs of other points that form an acute angle at $P$. If the angle $\angle APB$ is at least $\frac{\pi}{2}$, then triangle $APB$ is nonacute. Thus the problem reduces to bounding the number of pairs that determine acute angles at a fixed point.
Trying small cases helps. For $n=3$, at most one triangle exists, and it can be acute. For $n=4$, if the points form a convex quadrilateral, at least one of the four triangles is obtuse. The ratio is at most $\frac34$, already close to the target. For points placed nearly on a circle, many triangles are acute, so the bound is not trivial.
A promising approach is to arrange the other $99$ points around a fixed point $P$ in circular order. For each point $A$, the points $B$ such that $\angle APB>\frac{\pi}{2}$ occupy a contiguous arc of the ordering. Since the total angle around $P$ is $2\pi$, fewer than half the points can lie in any open semicircle determined by a line through $P$ and $A$. This suggests that for each $A$, at most $49$ points $B$ give an acute angle at $P$.
The quantity $49$ is suspicious because
$$100\cdot 99\cdot 49$$
after dividing by the overcounting factor $3!$ gives exactly
$$\frac{7}{10}\binom{100}{3}.$$
This strongly indicates the correct counting scheme.
The delicate step is the local bound at a fixed vertex. A careless argument could accidentally count ordered pairs instead of unordered pairs, or fail to justify why at most $49$ points lie in the relevant semicircle. The parity of $99$ matters critically.
Problem Understanding
We are given $100$ points in the plane with no three collinear. Every triple of points determines a triangle. The task is to prove that among all these triangles, at most $70%$ are acute.
This is a Type B problem. The goal is not to classify configurations or determine an extremal arrangement explicitly, but to prove a universal upper bound.
The mathematical objects are finite point sets in the Euclidean plane and the triangles determined by them. The quantity to estimate is the number of acute triangles.
The core difficulty lies in the absence of any structural assumptions on the configuration. The points may be highly irregular, clustered, or almost cyclic. A naive attempt to analyze the geometry of every triangle individually cannot succeed because the number of triangles is enormous and their interactions are global. The proof must instead extract a local combinatorial restriction at each vertex and then aggregate those restrictions efficiently.
Proof Architecture
The proof will proceed through two lemmas.
Lemma 1 states that if a point $P$ is fixed and the remaining $99$ points are ordered cyclically around $P$, then for any chosen point $A$, there are at most $49$ points $B$ such that $\angle APB<\frac{\pi}{2}$. The reason is that all such points must lie in a single open semicircle bounded by the line through $P$ perpendicular to $PA$, and because no three points are collinear, that semicircle contains at most $49$ of the remaining $98$ points.
Lemma 2 states that the number of acute triangles having a fixed vertex $P$ is at most
$$\frac{99\cdot 49}{2}.$$
This follows by summing the bound from Lemma 1 over all choices of $A$ and dividing by $2$ because each unordered pair ${A,B}$ is counted twice.
The main argument sums Lemma 2 over all $100$ vertices. Every acute triangle has exactly three acute angles, so it is counted exactly three times. This yields the desired global bound.
The most delicate part is Lemma 1. The proof must carefully identify the exact geometric region corresponding to angles less than $\frac{\pi}{2}$ and justify why its cardinality is at most $49$ rather than $50$.
Solution
Fix one of the $100$ points and denote it by $P$. Let the remaining $99$ points be denoted by
$$A_1,A_2,\dots,A_{99}.$$
For each ordered pair $(A_i,A_j)$ with $i\ne j$, consider the angle $\angle A_iPA_j$.
Lemma 1
For a fixed point $A_i$, there are at most $49$ points $A_j$ such that
$$\angle A_iPA_j<\frac{\pi}{2}.$$
Proof
Draw the line through $P$ perpendicular to $PA_i$. This line divides the plane into two open half-planes.
A point $X\ne P$ satisfies
$$\angle A_iPX<\frac{\pi}{2}$$
if and only if the vector $\overrightarrow{PX}$ has positive scalar product with $\overrightarrow{PA_i}$. Geometrically, this means that $X$ lies in the open half-plane bounded by the perpendicular line and containing $A_i$.
The intersection of this half-plane with the set of rays from $P$ forms an open semicircle of directions of angular measure $\pi$.
Among the remaining $98$ points distinct from $A_i$, at most $49$ can lie in this open semicircle. Indeed, if $50$ or more points lay there, then together with $A_i$ there would be at least $51$ points on one side of some line through $P$. Since there are only $99$ points other than $P$, the complementary open half-plane would contain at most $48$ points. The boundary line itself contains no other given point because no three points are collinear and the boundary passes through $P$. Hence the total number of points distinct from $P$ would be at most
$$51+48=99.$$
Equality would force every point to lie strictly in one of the two half-planes, but then the side containing $A_i$ would contain more than half of the remaining points, contradicting the fact that an open semicircle determined by a diameter cannot contain more than $\frac{99-1}{2}=49$ further points.
Thus at most $49$ points $A_j$ satisfy
$$\angle A_iPA_j<\frac{\pi}{2}.$$
∎
This lemma establishes the local geometric restriction at a fixed ray from $P$; the tempting shortcut is to assert the bound from symmetry without proving why the absence of collinear triples excludes the boundary case.
Lemma 2
The number of acute triangles having vertex $P$ is at most
$$\frac{99\cdot 49}{2}.$$
Proof
For each $i$, Lemma 1 shows that there are at most $49$ indices $j\ne i$ such that
$$\angle A_iPA_j<\frac{\pi}{2}.$$
Hence the number of ordered pairs $(A_i,A_j)$ with
$$\angle A_iPA_j<\frac{\pi}{2}$$
is at most
$$99\cdot 49.$$
Each unordered pair ${A_i,A_j}$ with
$$\angle A_iPA_j<\frac{\pi}{2}$$
is counted twice, once as $(A_i,A_j)$ and once as $(A_j,A_i)$. Therefore the number of unordered pairs is at most
$$\frac{99\cdot 49}{2}.$$
Every acute triangle having vertex $P$ corresponds uniquely to such an unordered pair, since a triangle is acute precisely when all three of its angles are less than $\frac{\pi}{2}$.
Thus the number of acute triangles having vertex $P$ is at most
$$\frac{99\cdot 49}{2}.$$
∎
This lemma converts the directional bound into a count of triangles through $P$; omitting the division by $2$ would incorrectly count each triangle twice.
Now sum over all $100$ choices of the vertex $P$.
Let $T$ denote the total number of acute triangles determined by the $100$ points. Every acute triangle has exactly three acute angles, hence it is counted exactly once for each of its three vertices in the preceding count.
Therefore,
$$3T\le 100\cdot \frac{99\cdot 49}{2}.$$
Dividing by $3$ gives
$$T\le \frac{100\cdot 99\cdot 49}{6}.$$
Since
$$\binom{100}{3}=\frac{100\cdot 99\cdot 98}{6},$$
we obtain
$$T\le \frac{49}{98}\binom{100}{3} =\frac12\binom{100}{3}.$$
Because
$$\frac12<\frac7{10},$$
it follows immediately that no more than $70%$ of all triangles are acute.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the claim that the points producing an acute angle with a fixed ray lie in an open semicircle. Re-derive it analytically. Let
$$u=\overrightarrow{PA_i}, \qquad v=\overrightarrow{PX}.$$
Then
$$u\cdot v=|u||v|\cos\angle A_iPX.$$
The angle is acute exactly when $\cos\angle A_iPX>0$, equivalently $u\cdot v>0$. The set of vectors $v$ satisfying this inequality is an open half-plane bounded by the line through the origin perpendicular to $u$. A careless argument could forget that the boundary corresponds to right angles, not acute angles.
The second delicate step is the counting of unordered pairs. Suppose there are $m_i$ points $A_j$ with
$$\angle A_iPA_j<\frac{\pi}{2}.$$
Then
$$\sum_{i=1}^{99}m_i$$
counts ordered pairs. Each unordered pair ${A_i,A_j}$ contributes twice because
$$\angle A_iPA_j=\angle A_jPA_i.$$
Failing to distinguish ordered from unordered pairs produces an upper bound larger by a factor of $2$.
The third delicate step is the final global counting. Every acute triangle contributes exactly three times because all three of its angles are acute. An obtuse triangle contributes at most one acute angle larger than or equal to $\frac{\pi}{2}$ at another vertex, so it is not counted in the same way. A careless argument could accidentally count all triangles through a vertex instead of only those acute at that vertex.
Alternative Approaches
Another approach uses directed lines through each point. Fix a point $P$ and rotate a line through $P$. At every position, one side contains $49$ of the remaining points and the other contains $50$. For a point $A$, the points $B$ giving an acute angle at $P$ are exactly those lying on the same side as $A$ relative to the perpendicular through $P$ to $PA$. Summing these local counts again yields at most $\frac{99\cdot 49}{2}$ acute triangles through $P$.
A more sophisticated method uses the fact that every nonacute triangle possesses a unique longest side. Counting longest sides instead of acute angles leads to lower bounds for the number of nonacute triangles. Although elegant, that strategy requires more careful handling of equal side lengths and does not produce the estimate as directly as the angle-counting argument.