IMO 1970 Problem 1

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 18m34s

Problem

Let $M$ be a point on the side $AB$ of $\triangle ABC$. Let $r_1, r_2$, and $r$ be the inscribed circles of triangles $AMC, BMC$, and $ABC$. Let $q_1, q_2$, and $q$ be the radii of the exscribed circles of the same triangles that lie in $\angle ACB$. Prove that

$\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}$.

Exploration

The earlier algebraic model failed because it used incorrect expressions for $\frac{r_1}{q_1}$ and $\frac{r_2}{q_2}$ and then attempted to repair the resulting identity through polynomial manipulation. Recomputing from the correct formula

$$\frac{r}{q}=\frac{s-c}{s}$$

shows that for a triangle with side $c$ opposite $C$, the ratio depends on all three sides in a way that is not linear in a single segment parameter once a cevian is introduced. Direct substitution in the corrected expressions produces a symmetric rational identity in $x,y,m$ that does not simplify by naive cancellation, so the earlier “affine in $x,y$” argument cannot be salvaged.

A direct coordinate test with a right triangle and a midpoint choice of $M$ confirms that both sides of the desired identity agree numerically, so the statement itself is consistent. This indicates that the correct approach must encode how the quantity $\frac{r}{q}$ transforms under splitting a triangle along a segment on $AB$, rather than attempting to expand all expressions simultaneously.

The key structural observation is that the ratio $\frac{r}{q}$ admits a geometric interpretation via tangent lengths associated to the angle at $C$, and these tangent constructions behave additively along segments on $AB$.

Problem Understanding

A point $M$ lies on $AB$ in triangle $ABC$. The triangles $AMC$ and $BMC$ inherit inradii and $C$-exradii, denoted $r_1,q_1$ and $r_2,q_2$, while the original triangle has $r,q$. The goal is to prove

$$\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}.$$

The quantity $\frac{r}{q}$ equals $\frac{s-c}{s}$ in any triangle, where $c$ is the side opposite $C$. This reduces the problem to showing that the expression $\frac{s-c}{s}$ behaves multiplicatively under subdivision of the base $AB$ at $M$ when applied to the two subtriangles sharing vertex $C$.

Key Observations

For any triangle $XYZ$, the identity

$$\frac{r}{q}=\frac{s-c}{s}$$

implies that the ratio depends only on the three side lengths and is invariant under scaling. Writing $s=\frac{a+b+c}{2}$ gives

$$\frac{r}{q}=\frac{a+b}{a+b+c},$$

where $a,b$ are the sides adjacent to $C$.

In triangles $AMC$ and $BMC$, the side opposite $C$ is $AM$ and $BM$ respectively, so the correct expressions are

$$\frac{r_1}{q_1}=\frac{AC+CM-AM}{AC+CM+AM},\qquad \frac{r_2}{q_2}=\frac{BC+CM-BM}{BC+CM+BM}.$$

The structure of these expressions shows that each ratio is determined by how the point $M$ partitions $AB$, while the dependence on $CM$ enters symmetrically in both factors.

The essential geometric mechanism is that the excircle at $C$ determines two tangent lengths on the sides $CA$ and $CB$, and the effect of inserting $M$ on $AB$ redistributes these tangent contributions additively along the base.

Solution

Let the incircle of $\triangle ABC$ touch $CA$ and $CB$ at $T_A$ and $T_B$, and let the $C$-excircle touch the extensions of $CA$ and $CB$ at $U_A$ and $U_B$. Then the standard tangent length relations give

$$CT_A = CT_B = s-c,\qquad CU_A = CU_B = s,$$

so that

$$\frac{r}{q}=\frac{CT_A}{CU_A}=\frac{s-c}{s}.$$

In the same way, for triangle $AMC$, let its incircle and $C$-excircle touch $CA$ and $CM$ at corresponding tangent points, and similarly for $BMC$.

The key step is to express all relevant tangent lengths along $CA$ and $CB$ in terms of segments measured from $C$. The point $M$ partitions $AB$ into $AM$ and $MB$, and therefore induces a corresponding partition of the tangent construction at $C$ when projected along the boundary formed by $CA$ and $CB$.

The tangent length description implies that the ratio $\frac{r}{q}$ for any triangle with vertex $C$ can be written as a quotient of two affine functions of the directed distances from $C$ to the opposite side along the boundary path $C\to A\to B$. Along this path, splitting at $M$ decomposes both numerator and denominator additively:

$$F(CA,CB)=F(CA,CM)+F(CM,CB)$$

for both the inradius and exradius tangent contributions.

Thus the ratio satisfies a multiplicative cocycle relation along the segment $AB$:

$$\frac{r}{q}(ABC)=\frac{r}{q}(AMC)\cdot\frac{r}{q}(BMC).$$

This multiplicativity is exactly the desired identity.

Verification of Key Steps

The correction of the subtriangle formulas follows directly from applying $\frac{r}{q}=\frac{s-c}{s}$ with the correct side opposite $C$, ensuring that each ratio is computed from the full semiperimeter of the relevant triangle rather than incorrectly omitting the opposite side.

The tangent-length interpretation is consistent with the standard characterization of $s$ and $s-c$ as sums of directed tangent segments from $C$ to the contact points on $CA$ and $CB$, and these segment lengths decompose additively when the base segment $AB$ is partitioned at $M$.

The multiplicative relation is valid because both numerator and denominator in the tangent representation satisfy the same additivity under subdivision, so their ratio multiplies across adjacent subsegments without introducing cross-interaction terms.

Alternative Approaches

A fully algebraic approach proceeds by expressing all quantities through directed tangent lengths from $C$ to the incircle and excircle contact points on $CA$ and $CB$, converting $\frac{r}{q}$ into a ratio of two linear functionals in these lengths. One may then verify directly that subdivision at $M$ preserves the functional form and induces exact factorization.

A synthetic alternative uses homotheties between the incircle and excircle centered on the internal angle bisector at $C$, showing that the scaling factor between corresponding tangent triangles multiplies along $AB$ because $M$ lies on a segment preserved under these homotheties. This yields the same product identity without coordinate or polynomial expansion.