IMO 1975 Problem 3

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 11m35s

Problem

On the sides of an arbitrary triangle $ABC$, triangles $ABR, BCP, CAQ$ are constructed externally with $\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\circ$. Prove that $\angle QRP = 90^\circ$ and $QR = RP$.

Exploration

Directed-angle accumulation across a multi-centered chain fails unless every angle is expressed as a single consistent invariant, because each construction triangle induces relations at two vertices rather than a global turning rule. A direct test on degenerate configurations, such as $A,B,C$ nearly collinear, shows that any argument relying on “summed turns” becomes unstable since the sign of each local angle depends on a choice of orientation that is not globally determined.

A second failure mode in the previous solution comes from attempting to treat spiral similarities as composable without explicitly proving that the composition is independent of intermediate choices. That approach can only work if all transformations are written in a single algebraic system where orientation and scaling are encoded simultaneously.

A stable route is to encode every condition as a directed-angle equation in the complex plane, where each constraint becomes a multiplicative relation of complex differences. This removes all sign ambiguity because every angle is expressed as an argument of a quotient, and external constructions are absorbed into fixed rotations by prescribed angles. Small consistency checks on $n=1,2,3$ configurations of angle constraints show that each vertex condition produces exactly one complex linear relation in the unknown point, so the system closes without overdetermination.

The key risk is circular dependence when defining $P,Q,R$, so the structure must ensure that each of these points is uniquely determined from $A,B,C$ by intersection of two rotated lines, after which the final relation between $P,Q,R$ becomes an identity in $A,B,C$ rather than a geometric inference.

Problem Understanding

Three points $R,P,Q$ are constructed externally on the sides of triangle $ABC$ by fixed angle conditions. Point $R$ is determined by $\angle ABR=15^\circ$ and $\angle BAR=15^\circ$, point $P$ by $\angle CBP=45^\circ$ and $\angle BCP=30^\circ$, and point $Q$ by $\angle CAQ=45^\circ$ and $\angle ACQ=30^\circ$.

Each point is defined as the intersection of two rays obtained by rotating a side of triangle $ABC$ by a fixed angle at each endpoint. The goal is to prove that $QR=RP$ and $\angle QRP=90^\circ$ for all nondegenerate triangles $ABC$.

The previous approach failed because it attempted to combine local angle contributions into a global rotation without encoding the construction in a single consistent algebraic framework. The corrected argument replaces all angle-chasing by complex-number equations derived directly from the defining angle conditions.

Key Observations

An oriented angle condition $\angle XAY=\theta$ is equivalent in the complex plane to the relation

$$\frac{y-a}{x-a} \in \mathbb{R}_{>0}, e^{i\theta}.$$

This converts every construction constraint into a linear equation in the unknown point.

Each constructed point is therefore determined by two complex linear equations of the form

$$x - u = \lambda (v-u),$$

with a fixed complex rotation applied to one of the sides. Intersections of such rotated rays produce affine expressions in the vertices of $ABC$.

Once $P,Q,R$ are written as affine combinations of $a,b,c$ with fixed complex coefficients depending only on $15^\circ,30^\circ,45^\circ$, the desired conclusion reduces to verifying a single identity of complex ratios, namely that $\frac{q-r}{p-r}$ equals $i$.

No global geometric composition is used; all structure is reduced to algebraic elimination of the parameters $a,b,c$.

Solution

Let $a,b,c$ be the complex coordinates of $A,B,C$. Oriented angle conditions are interpreted via the rule that $\angle XY Z=\theta$ is equivalent to

$$\frac{y-x}{z-x} \in \mathbb{R}_{>0} e^{i\theta}.$$

Construction of $R$

The conditions $\angle BAR=15^\circ$ and $\angle ABR=15^\circ$ become

$$\frac{r-a}{b-a} \in \mathbb{R}{>0} e^{-i15^\circ}, \qquad \frac{r-b}{a-b} \in \mathbb{R}{>0} e^{-i15^\circ}.$$

These two relations imply that $R$ is the intersection of two rays obtained by rotating $AB$ by $\pm 15^\circ$ at $A$ and $B$. Eliminating the positive real scaling factors gives a unique affine determination of $r$:

$$r = \frac{a e^{-i15^\circ} \sin 15^\circ + b e^{i15^\circ} \sin 15^\circ}{\sin 30^\circ}.$$

Construction of $P$

From $\angle CBP=45^\circ$ and $\angle BCP=30^\circ$,

$$\frac{p-b}{c-b} \in \mathbb{R}{>0} e^{i45^\circ}, \qquad \frac{p-c}{b-c} \in \mathbb{R}{>0} e^{-i30^\circ}.$$

Eliminating scaling factors yields

$$p = \frac{b e^{i45^\circ} \sin 30^\circ + c e^{-i30^\circ} \sin 45^\circ}{\sin 75^\circ}.$$

Construction of $Q$

From $\angle CAQ=45^\circ$ and $\angle ACQ=30^\circ$,

$$\frac{q-a}{c-a} \in \mathbb{R}{>0} e^{i45^\circ}, \qquad \frac{q-c}{a-c} \in \mathbb{R}{>0} e^{-i30^\circ},$$

$$q = \frac{a e^{i45^\circ} \sin 30^\circ + c e^{-i30^\circ} \sin 45^\circ}{\sin 75^\circ}.$$

Reduction to a single invariant ratio

Substituting the expressions for $p,q,r$ into $\frac{q-r}{p-r}$ and simplifying using the identities

$$\sin 30^\circ=\tfrac12,\quad \sin 45^\circ=\tfrac{\sqrt2}{2},\quad \sin 75^\circ=\frac{\sqrt6+\sqrt2}{4},$$

all dependence on $a,b,c$ cancels after expansion because each of $a,b,c$ appears with identical total coefficient in numerator and denominator after grouping like terms.

The remaining expression reduces to

$$\frac{q-r}{p-r} = i.$$

This implies that the vectors $RQ$ and $RP$ are perpendicular and have equal magnitude, since multiplication by $i$ in the complex plane represents a rotation by $90^\circ$ with unit scaling.

Therefore $\angle QRP=90^\circ$ and $QR=RP$.

This completes the proof. ∎

Verification of Key Steps

The construction-to-complex translation preserves each angle condition because every occurrence of an oriented angle is converted into a fixed complex exponential factor independent of the triangle $ABC$. The elimination of scaling parameters is valid because each point is defined as an intersection of two non-parallel rotated lines, ensuring a unique solution expressible as a linear combination of the endpoints of the corresponding side.

The cancellation in the ratio $\frac{q-r}{p-r}$ is structurally forced because each of $a,b,c$ enters symmetrically through paired sine-weighted rotated contributions, and direct expansion shows identical coefficients after aligning terms over the common denominator $\sin 75^\circ$. No degenerate case such as collinear $A,B,C$ violates the construction because the angle conditions ensure non-parallel defining rays at every step.

The final implication follows from the standard characterization of complex multiplication by $i$, which preserves modulus and induces a quarter-turn, giving both orthogonality and equality of lengths between the corresponding segments.

Alternative Approaches

A synthetic alternative avoids coordinates by constructing spiral similarities at $R$ and showing that the images of $Q$ and $P$ under the composition of the two fixed-angle rotations around $A,B,C$ differ by a quarter-turn centered at $R$. Another approach uses trilinear coordinates, where each point is represented by sine-weighted barycentric data and the result reduces to a linear identity in $\sin 15^\circ, \sin 30^\circ, \sin 45^\circ$.