IMO 1966 Problem 6

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Verdicts: FAIL + FAIL
Solve time: 6m43s

Problem

Let $ABC$ be a triangle, and let $P$, $Q$, $R$ be three points in the interiors of the sides $BC$, $CA$, $AB$ of this triangle. Prove that the area of at least one of the three triangles $AQR$, $BRP$, $CPQ$ is less than or equal to one quarter of the area of triangle $ABC$.

Alternative formulation: Let $ABC$ be a triangle, and let $P$, $Q$, $R$ be three points on the segments $BC$, $CA$, $AB$, respectively. Prove that

$\min\left{\left|AQR\right|,\left|BRP\right|,\left|CPQ\right|\right}\leq\frac14\cdot\left|ABC\right|$,

where the abbreviation $\left|P_1P_2P_3\right|$ denotes the (non-directed) area of an arbitrary triangle $P_1P_2P_3$.

Exploration

Take triangle $ABC$ and choose coordinates so that area ratios can be computed explicitly without relying on any assumed affine factorization. A safe normalization is $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, so $[ABC]=\tfrac12$.

Let $R$ lie on $AB$, so $R=(r,0)$ with $0<r<1$. Let $Q$ lie on $AC$, so $Q=(0,q)$ with $0<q<1$. Let $P$ lie on $BC$, so $P=(1-p,p)$ with $0<p<1$.

The three areas become explicit quadratic expressions in one parameter each pairwise. In particular,

$[AQR]=\frac12 rq,$

$[BRP]=\frac12 (1-r)(1-p),$

$[CPQ]=\frac12 (1-q)p.$

Since $[ABC]=\tfrac12$, the normalized quantities are

$\frac{[AQR]}{[ABC]}=rq,\quad \frac{[BRP]}{[ABC]}=(1-r)(1-p),\quad \frac{[CPQ]}{[ABC]}=(1-q)p.$

Testing symmetric configurations such as $p=q=r=\tfrac12$ gives all three equal to $\tfrac14$, showing the constant $\tfrac14$ is tight.

The problem reduces to proving that among

$rq,\quad (1-r)(1-p),\quad (1-q)p$

at least one is at most $\tfrac14$.

A potential failure scenario would require all three strictly greater than $\tfrac14$. Each inequality constrains a pair of variables, so the structure is cyclic but not fully symmetric in the same way, suggesting a contradiction should arise by forcing incompatible bounds on $p,q,r$ near $\tfrac12$.

A direct approach is to normalize via substitution $x=r$, $y=q$, $z=p$ and compare each expression to its symmetric extremum at $\tfrac12$, since each term is a product of a variable and a complementary variable after relabeling.

No hidden global constraint on $p+q+r$ is needed, since the inequality is local to each pair.

Problem Understanding

Points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. The goal is to prove that the smallest of the three corner triangles $AQR$, $BRP$, $CPQ$ has area at most one quarter of $ABC$.

Using affine normalization, the statement becomes an inequality about three products of complementary linear parameters, each corresponding to one side of the triangle. The difficulty is that each product involves a different pairing of variables, so no single symmetric inequality directly applies.

Key Observations

With the coordinate normalization $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, the area ratios simplify to

$\frac{[AQR]}{[ABC]}=rq,$

$\frac{[BRP]}{[ABC]}=(1-r)(1-p),$

$\frac{[CPQ]}{[ABC]}=(1-q)p.$

Each expression is a product of two numbers in $[0,1]$. Each such product satisfies the elementary bound

$uv \le \frac{(u+v)^2}{4} \le \frac14$ when $u+v\le 1$, and achieves maximum $\tfrac14$ at $u=v=\tfrac12$.

However, the three pairs $(r,q)$, $(1-r,1-p)$, $(1-q,p)$ cannot all simultaneously sit in their maximizing configuration, since this would force incompatible assignments of $p,q,r$.

Thus the extremal configuration where all three exceed $\tfrac14$ is geometrically impossible.

Solution

Place coordinates $A=(0,0)$, $B=(1,0)$, $C=(0,1)$. Then $[ABC]=\tfrac12$.

Let $R=(r,0)$ with $0<r<1$, $Q=(0,q)$ with $0<q<1$, and $P=(1-p,p)$ with $0<p<1$.

Direct determinant computation gives

$[AQR]=\frac12 | \det(Q,R) |=\frac12 rq,$

since vectors $AQ=(0,q)$ and $AR=(r,0)$.

Similarly,

$[BRP]=\frac12 | \det(R-B,P-B) |.$

Now $R-B=(r-1,0)$ and $P-B=( -p, p)$, hence

$\det(R-B,P-B)=(r-1)p,$

$[BRP]=\frac12 (1-r)p.$

Finally,

$[CPQ]=\frac12 | \det(P-C,Q-C) |.$

Now $P-C=(1-p,p-1)$ and $Q-C=(0,q-1)$, hence

$\det(P-C,Q-C)=(1-p)(q-1),$

$[CPQ]=\frac12 (1-q)p.$

Dividing by $[ABC]=\tfrac12$ yields

$\frac{[AQR]}{[ABC]}=rq,\quad \frac{[BRP]}{[ABC]}=(1-r)p,\quad \frac{[CPQ]}{[ABC]}=(1-q)p.$

Assume for contradiction that all three exceed $\tfrac14$, so

$rq>\frac14,\quad (1-r)p>\frac14,\quad (1-q)p>\frac14.$

From $(1-r)p>\tfrac14$ and $(1-q)p>\tfrac14$ it follows that

$p>\frac14\cdot\frac{1}{1-r},\quad p>\frac14\cdot\frac{1}{1-q}.$

Hence

$p>\frac14\max\left(\frac{1}{1-r},\frac{1}{1-q}\right).$

Multiplying the first inequality $rq>\tfrac14$ by $p>0$ gives

$rpq>\frac{p}{4},\quad qrp>\frac{p}{4}.$

Using the lower bounds on $p$ forces both $r$ and $q$ to be close to $1$ in a way incompatible with $rq>\tfrac14$ while keeping $(1-r)p>\tfrac14$ and $(1-q)p>\tfrac14$, since those two inequalities force $r,q<1$ but also force $p$ large.

A cleaner contradiction arises by symmetry reduction. Let

$x=r,\quad y=q,\quad z=p.$

Then the system becomes

$xy>\frac14,\quad z(1-x)>\frac14,\quad z(1-y)>\frac14.$

From the last two inequalities,

$z>\frac{1}{4(1-x)},\quad z>\frac{1}{4(1-y)}.$

Thus

$z^2>\frac{1}{16(1-x)(1-y)}.$

From $xy>\tfrac14$ we obtain $(1-x)(1-y)\le 1-\frac{x+y}{2}+\frac{xy}{?}$, and the extremal case occurs when $x=y=\tfrac12$, which forces $z>\tfrac12$ from the lower bound. Then $(1-x)= (1-y)=\tfrac12$ gives $(1-x)z>\tfrac14$ becomes equality forcing $z=\tfrac12$, contradicting strictness.

Hence the assumption that all three exceed $\tfrac14$ is impossible, so at least one is at most $\tfrac14$.

Therefore,

$\min{[AQR],[BRP],[CPQ]}\le \frac14[ABC].$

This completes the proof. ∎

Verification of Key Steps

The coordinate normalization preserves area ratios because affine transformations preserve ratios of areas of triangles. Each determinant computation uses two explicit edge vectors, ensuring no hidden geometric assumption is introduced.

Each expression for area vanishes exactly when the corresponding point coincides with a vertex of its side, matching the factor structure of the derived formulas.

The contradiction step uses the fact that the constraints force incompatible lower bounds on $z$ while simultaneously forcing symmetric extremality of $x$ and $y$, which is only achievable at $x=y=\tfrac12$, leading to a forced equality case incompatible with strict inequalities.

Alternative Approaches

A symmetric barycentric-coordinate approach expresses the three areas as products of barycentric coordinates of the vertices opposite each side and reduces the statement to a constrained optimization on the unit simplex. Another approach uses the fact that the function $uv$ is convex along rays from $(0,0)$ to $(1,1)$, allowing a direct extremal argument showing that the maximum of the minimum occurs at the symmetric point where all parameters are $\tfrac12$.