IMO 1964 Problem 3
Let $\triangle ABC$ have semiperimeter $s$ and area $\Delta$, with inradius $r$.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 10m26s
Problem
A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from $\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).
Exploration
Let $\triangle ABC$ have semiperimeter $s$ and area $\Delta$, with inradius $r$. The configuration introduces three lines, each parallel to one side of the triangle and tangent to the incircle. Each such line truncates a smaller triangle similar to the original triangle, leaving three corner triangles near the vertices.
A key geometric feature is that similarity appears repeatedly: a line parallel to a side produces a homothetic copy of the full triangle at the opposite vertex. This suggests that the radii of the inscribed circles in the small triangles scale linearly with the corresponding similarity ratios.
The central incircle is known to have radius $r=\frac{\Delta}{s}$, and triangle heights satisfy $h_a=\frac{2\Delta}{a}$, with analogous expressions for $h_b,h_c$. The distance from each side to the corresponding tangent line parallel to it must equal $2r$, since the incircle lies at distance $r$ from each side and the tangent is shifted by another $r$.
This leads to similarity ratios of the form $1-\frac{a}{s}$ and cyclic variants. The main difficulty lies in correctly summing the squared similarity ratios and expressing everything symmetrically in $a,b,c$ without introducing asymmetric intermediate quantities.
The decisive step is reducing the total area to a symmetric polynomial in $a,b,c$ divided by a power of $s$.
Problem Understanding
The problem asks for the total area of four circles, one inscribed in the original triangle and one in each of the three smaller triangles formed by drawing tangents to the incircle that are parallel to the sides of the triangle.
This is a determination problem, so it is Type C. The expected structure is that each of the four circles has area proportional to the square of a corresponding inradius, and all inradii arise from similarity with the original triangle.
The key challenge is to compute the inradii of the three truncated corner triangles in terms of $a,b,c$, then sum their areas with the area of the original incircle.
The final expression depends only on symmetric functions of $a,b,c$ and the semiperimeter $s$.
The final result is
$$\boxed{\frac{\pi \Delta^2 (a^2+b^2+c^2)}{s^4}}$$
where $s=\frac{a+b+c}{2}$ and $\Delta$ is the area of $\triangle ABC$.
Proof Architecture
The first lemma identifies the distance between each side of the triangle and the corresponding tangent line parallel to that side, expressed in terms of the inradius $r$.
The second lemma establishes that each truncated corner triangle is similar to $\triangle ABC$ and computes its similarity ratio as $1-\frac{a}{s}$ and cyclic permutations.
The third lemma computes the inradius of a triangle under similarity scaling.
The fourth lemma performs the algebraic simplification of the sum of squared similarity ratios into a symmetric polynomial in $a,b,c$ and $s$.
The main argument assembles these results to compute the sum of the four areas.
The most delicate point is the computation of the shift distance $2r$, since an incorrect geometric offset breaks all subsequent scaling relations.
Solution
Lemma 1
The distance between each side of $\triangle ABC$ and the corresponding tangent line to the incircle that is parallel to that side equals $2r$.
In the triangle, the incircle is tangent to each side, hence its center $I$ is at distance $r$ from each side. Consider the side $BC$. The line parallel to $BC$ and tangent to the incircle on the side of vertex $A$ must be at distance $r$ from $I$ in the direction perpendicular to $BC$ and on the same side as $A$. Since $I$ itself is at distance $r$ from $BC$, the total distance from $BC$ to this tangent line equals $r+r=2r$.
The same reasoning applies cyclically to the other two sides, since the incircle is equidistant from all sides.
This establishes that each truncating line is obtained from the corresponding side by a perpendicular translation of length $2r$ toward the opposite vertex, and any shortcut treating the shift as $r$ would contradict the definition of tangency to the incircle.
Lemma 2
Let the line parallel to $BC$ tangent to the incircle on the side of $A$ intersect $AB$ and $AC$ at points $B_1$ and $C_1$. Then $\triangle AB_1C_1$ is similar to $\triangle ABC$, and its similarity ratio equals $1-\frac{a}{s}$. Analogous statements hold cyclically.
Since $B_1C_1 \parallel BC$, the triangles $\triangle AB_1C_1$ and $\triangle ABC$ are similar by corresponding angle equality. Let $h_a$ be the altitude from $A$ to $BC$. The altitude of $\triangle AB_1C_1$ from $A$ to $B_1C_1$ equals $h_a-2r$ by Lemma 1.
Thus the similarity ratio $k_a$ satisfies
$$k_a=\frac{h_a-2r}{h_a}=1-\frac{2r}{h_a}.$$
Using $h_a=\frac{2\Delta}{a}$ and $r=\frac{\Delta}{s}$, one obtains
$$\frac{2r}{h_a}=\frac{2(\Delta/s)}{2\Delta/a}=\frac{a}{s},$$
hence
$$k_a=1-\frac{a}{s}=\frac{s-a}{s}.$$
The same computation holds for the other two vertices by cyclic permutation of $a,b,c$. Any attempt to assign different similarity factors to different vertices fails because both $r$ and $\Delta$ are global invariants of the same triangle.
Lemma 3
If a triangle is similar to another triangle with similarity ratio $k$, then their inradii are in the same ratio.
Under a similarity transformation with ratio $k$, all lengths scale by $k$, and distances from corresponding sides scale by $k$ as well. The inradius is defined as the distance from the incenter to each side, hence it also scales by $k$.
Therefore, if $r$ is the inradius of $\triangle ABC$, the inradius of the truncated triangle at $A$ equals $k_a r$, and similarly for the other two.
This rules out any nonlinear dependence of the small inradii on $k_a$, since any such dependence would contradict uniform scaling of perpendicular distances.
Lemma 4
The identity
$$(s-a)^2+(s-b)^2+(s-c)^2=a^2+b^2+c^2-s^2$$
holds, where $s=\frac{a+b+c}{2}$.
Expanding each term gives
$$(s-a)^2+(s-b)^2+(s-c)^2=3s^2-2s(a+b+c)+(a^2+b^2+c^2).$$
Since $a+b+c=2s$, substitution yields
$$3s^2-4s^2+(a^2+b^2+c^2)=a^2+b^2+c^2-s^2.$$
No alternative simplification avoids this cancellation, since the symmetry forces dependence only on $s$ and $a^2+b^2+c^2$.
Main Argument
The original incircle has area $\pi r^2$.
Each truncated triangle is similar to $\triangle ABC$, hence its inradius equals $k_a r$, $k_b r$, $k_c r$ respectively. Their corresponding circle areas are
$$\pi k_a^2 r^2,\quad \pi k_b^2 r^2,\quad \pi k_c^2 r^2.$$
The total area $T$ of the four circles is
$$T=\pi r^2\left(1+k_a^2+k_b^2+k_c^2\right).$$
Using $k_a=\frac{s-a}{s}$ and cyclic variants,
$$k_a^2+k_b^2+k_c^2=\frac{(s-a)^2+(s-b)^2+(s-c)^2}{s^2}.$$
Applying Lemma 4,
$$k_a^2+k_b^2+k_c^2=\frac{a^2+b^2+c^2-s^2}{s^2}.$$
Thus,
$$1+k_a^2+k_b^2+k_c^2=\frac{s^2+a^2+b^2+c^2-s^2}{s^2}=\frac{a^2+b^2+c^2}{s^2}.$$
Hence,
$$T=\pi r^2 \cdot \frac{a^2+b^2+c^2}{s^2}.$$
Substituting $r=\frac{\Delta}{s}$ yields
$$T=\pi \frac{\Delta^2}{s^2}\cdot \frac{a^2+b^2+c^2}{s^2} =\frac{\pi \Delta^2 (a^2+b^2+c^2)}{s^4}.$$
This completes the determination of the total area.
Verification of Key Steps
The first delicate point is the computation of the distance $2r$ between each side and the corresponding tangent line. Any mistaken assumption that the shift equals $r$ would incorrectly produce similarity ratios $1-\frac{a}{2s}$, leading to a non-symmetric final expression inconsistent with cyclic structure.
The second delicate point is the relation $h_a=\frac{2\Delta}{a}$. This connects metric quantities of the triangle to its area and ensures that the ratio $\frac{2r}{h_a}$ simplifies to $\frac{a}{s}$. Without this identity, the dependence on $a,b,c$ would not reduce to a symmetric polynomial.
The third delicate point is the algebraic cancellation in Lemma 4. A direct expansion without careful collection of $s$-terms can incorrectly suggest dependence on mixed products such as $ab$, but all such terms cancel due to the identity $a+b+c=2s$.
Alternative Approaches
One alternative approach replaces similarity arguments with affine transformations sending $\triangle ABC$ to an equilateral triangle, where the configuration becomes symmetric and the inradius relations are immediate. The result is then transferred back using invariance of ratios under affine maps.
Another approach uses barycentric coordinates to express the tangent lines explicitly and derive the side ratios of the truncated triangles directly. This method is more computational and obscures the geometric origin of the factor $2r$, but it leads to the same final symmetric expression after simplification.