IMO 1964 Problem 3
Let the inradius of $\triangle ABC$ be $r$, and let its area be $\Delta$.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 3m04s
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 the inradius of $\triangle ABC$ be $r$, and let its area be $\Delta$. The inscribed circle of the original triangle contributes area $\pi r^2$. The three smaller triangles cut off by the new tangents each have their own incircle radii, say $r_a,r_b,r_c$. The task is to compute
$$\pi(r^2+r_a^2+r_b^2+r_c^2).$$
A first guess is that the answer may simplify to a multiple of the area of the original triangle. Testing the equilateral case gives useful numerical evidence.
Take an equilateral triangle of side length $s$. Then
$$r=\frac{s\sqrt3}{6}.$$
The tangent parallel to a side lies halfway between that side and the opposite vertex because the center is equidistant from the two parallel tangents. Hence each corner triangle is similar to the original triangle with linear scale factor $1/2$. Therefore each smaller inradius equals $r/2$. The total area of the four circles becomes
$$\pi\left(r^2+3\left(\frac r2\right)^2\right) = \frac74\pi r^2.$$
Since
$$r^2=\frac{s^2}{12},$$
the total is
$$\frac{7\pi s^2}{48}.$$
Because the original triangle area is
$$\Delta=\frac{\sqrt3 s^2}{4},$$
the ratio to $\Delta$ is
$$\frac{7\pi}{12\sqrt3}.$$
Nothing especially natural emerges, so the answer probably depends on $a,b,c$ through a symmetric quadratic expression.
A more structural approach is needed. The corner triangle near $A$ is formed by side lines $AB,AC$ and the tangent parallel to $BC$. Hence it is similar to $\triangle ABC$. Determining its similarity ratio becomes the central task.
Suppose the tangent parallel to $BC$ touches the incircle at the point opposite $BC$. The distance between $BC$ and this new tangent equals $2r$, because both are tangents to the same circle on opposite sides through the center. Since the altitude from $A$ to $BC$ equals
$$h_a=\frac{2\Delta}{a},$$
the altitude of the small triangle at $A$ equals
$$h_a-2r.$$
Because similar triangles scale all lengths equally, the similarity factor is
$$\frac{h_a-2r}{h_a}.$$
Using
$$\Delta=rs,\qquad s=\frac{a+b+c}{2},$$
gives
$$h_a=\frac{2rs}{a},$$
hence
$$\frac{h_a-2r}{h_a} = 1-\frac{a}{s} = \frac{s-a}{s}.$$
Thus
$$r_a=r\frac{s-a}{s}.$$
Cyclically,
$$r_b=r\frac{s-b}{s},\qquad r_c=r\frac{s-c}{s}.$$
Now compute
$$r^2+r_a^2+r_b^2+r_c^2 = r^2\left(1+\frac{(s-a)^2+(s-b)^2+(s-c)^2}{s^2}\right).$$
The algebra must simplify symmetrically. Since
$$s-a=\frac{b+c-a}{2},$$
and similarly for the others, expanding directly is possible. Another route is to use
$$(s-a)+(s-b)+(s-c)=s.$$
Compute
$$(s-a)^2+(s-b)^2+(s-c)^2 = a^2+b^2+c^2-s^2.$$
Substituting
$$s^2=\frac{(a+b+c)^2}{4}$$
should produce a clean form.
Carrying out the expansion,
$$a^2+b^2+c^2-s^2 = \frac{3(a^2+b^2+c^2)-2(ab+bc+ca)}{4}.$$
Then
$$1+\frac{a^2+b^2+c^2-s^2}{s^2} = \frac{4(a^2+b^2+c^2)}{(a+b+c)^2}.$$
Hence
$$r^2+r_a^2+r_b^2+r_c^2 = \frac{4r^2(a^2+b^2+c^2)}{(a+b+c)^2}.$$
Since
$$r=\frac{\Delta}{s}=\frac{2\Delta}{a+b+c},$$
the total circle area becomes
$$\frac{16\pi\Delta^2(a^2+b^2+c^2)}{(a+b+c)^4}.$$
Using Heron's formula,
$$16\Delta^2 = 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4.$$
This yields a fully side-length expression.
The step most likely to conceal an error is the derivation of the similarity ratio
$$\frac{s-a}{s}.$$
A sign mistake there would corrupt the entire computation.
Problem Understanding
The problem concerns a triangle $\triangle ABC$ with an incircle. Through the incircle, three additional tangents are drawn, each parallel to one side of the triangle. Each such tangent cuts off a smaller triangle at the opposite vertex. Each of these three smaller triangles has its own incircle.
The objective is to determine the sum of the areas of the four circles, namely the original incircle and the three incircles of the corner triangles, expressed entirely in terms of the side lengths $a,b,c$.
This is a Type C problem because a definite quantity must be determined exactly.
The main geometric difficulty is identifying the sizes of the three smaller incircles. A direct coordinate computation would become cumbersome because the positions of the tangency lines depend on the original incenter geometry. The crucial simplification is that each corner triangle is similar to the original triangle, so its inradius scales linearly with a similarity factor that can be computed from altitudes.
The answer should be
$$\boxed{ \frac{\pi(a^2+b^2+c^2)\bigl(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\bigr)} {(a+b+c)^4} }.$$
The expression is symmetric in $a,b,c$, which matches the symmetry of the construction.
Proof Architecture
The proof proceeds through four claims.
Claim 1
The triangle cut off near vertex $A$ by the tangent parallel to $BC$ is similar to $\triangle ABC$, with similarity ratio
$$\frac{s-a}{s},$$
where
$$s=\frac{a+b+c}{2}.$$
The proof uses the fact that corresponding altitudes in similar triangles scale equally, together with the relation $\Delta=rs$.
Claim 2
If $r$ is the inradius of $\triangle ABC$, then the inradius $r_a$ of the corner triangle near $A$ satisfies
$$r_a=r\frac{s-a}{s},$$
and cyclically for $r_b,r_c$.
This follows because inradii scale linearly under similarity.
Claim 3
The sum of the squares of the four radii satisfies
$$r^2+r_a^2+r_b^2+r_c^2 = \frac{4r^2(a^2+b^2+c^2)}{(a+b+c)^2}.$$
The proof is algebraic and depends on simplifying
$$(s-a)^2+(s-b)^2+(s-c)^2.$$
Claim 4
Using
$$r=\frac{2\Delta}{a+b+c}$$
and Heron's identity
$$16\Delta^2 = 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4,$$
the required total circle area equals the stated formula.
The most delicate step is Claim 1 because the distance between the two parallel tangents must be interpreted correctly as $2r$.
Solution
Let the inradius of $\triangle ABC$ be $r$, and let
$$s=\frac{a+b+c}{2}$$
be the semiperimeter. Denote the area of $\triangle ABC$ by $\Delta$.
For each side, draw the tangent to the incircle parallel to that side. Let the triangle cut off near vertex $A$ be denoted by $\triangle A_1$, and let its inradius be $r_a$. Define $\triangle B_1,\triangle C_1$ and radii $r_b,r_c$ analogously.
We compute
$$\pi(r^2+r_a^2+r_b^2+r_c^2).$$
Claim 1
The triangle $\triangle A_1$ is similar to $\triangle ABC$ with similarity ratio
$$\frac{s-a}{s}.$$
Proof
The side of $\triangle A_1$ opposite $A$ is the tangent parallel to $BC$. Hence the corresponding sides of $\triangle A_1$ and $\triangle ABC$ are parallel, so the two triangles are similar.
Let $h_a$ be the altitude of $\triangle ABC$ from $A$ to $BC$. Since
$$\Delta=\frac12 ah_a,$$
we obtain
$$h_a=\frac{2\Delta}{a}.$$
The line forming the base of $\triangle A_1$ is tangent to the incircle and parallel to $BC$. The distance between this tangent and the side $BC$ equals the distance between two parallel tangents to the same circle on opposite sides. Since the circle has radius $r$, this distance equals $2r$.
Therefore the altitude of $\triangle A_1$ equals
$$h_a-2r.$$
Since similar triangles have equal ratios of corresponding altitudes, the similarity factor is
$$\frac{h_a-2r}{h_a}.$$
Now use
$$\Delta=rs.$$
Then
$$h_a=\frac{2rs}{a},$$
hence
$$\frac{h_a-2r}{h_a} = \frac{\frac{2rs}{a}-2r}{\frac{2rs}{a}} = \frac{\frac{2r(s-a)}{a}}{\frac{2rs}{a}} = \frac{s-a}{s}.$$
Thus the similarity ratio is
$$\frac{s-a}{s}.$$
∎
This establishes the exact scaling factor of the corner triangle; replacing the altitude computation by an unproved proportionality argument would leave the central geometric step unjustified.
Claim 2
The inradii of the three corner triangles satisfy
$$r_a=r\frac{s-a}{s},\qquad r_b=r\frac{s-b}{s},\qquad r_c=r\frac{s-c}{s}.$$
Proof
Under a similarity transformation with ratio $k$, every length scales by the factor $k$. Since an inradius is a length, it scales linearly.
By Claim 1, $\triangle A_1$ is similar to $\triangle ABC$ with ratio
$$\frac{s-a}{s}.$$
Hence
$$r_a=r\frac{s-a}{s}.$$
The same argument applied cyclically yields
$$r_b=r\frac{s-b}{s},\qquad r_c=r\frac{s-c}{s}.$$
∎
This determines the three unknown radii exactly; omitting the similarity justification would make the formulas unsupported.
Claim 3
The sum of the squares of the four radii is
$$r^2+r_a^2+r_b^2+r_c^2 = \frac{4r^2(a^2+b^2+c^2)}{(a+b+c)^2}.$$
Proof
Substituting the formulas from Claim 2 gives
$$r^2+r_a^2+r_b^2+r_c^2 = r^2\left( 1+ \frac{(s-a)^2+(s-b)^2+(s-c)^2}{s^2} \right).$$
We simplify the numerator. Expanding,
$$\begin{aligned} &(s-a)^2+(s-b)^2+(s-c)^2 \ &= 3s^2-2s(a+b+c)+(a^2+b^2+c^2). \end{aligned}$$
Since
$$a+b+c=2s,$$
this becomes
$$a^2+b^2+c^2-s^2.$$
Therefore
$$\begin{aligned} r^2+r_a^2+r_b^2+r_c^2 &= r^2\left( 1+\frac{a^2+b^2+c^2-s^2}{s^2} \right) \ &= r^2\frac{a^2+b^2+c^2}{s^2}. \end{aligned}$$
Because
$$s=\frac{a+b+c}{2},$$
we have
$$s^2=\frac{(a+b+c)^2}{4}.$$
Hence
$$r^2+r_a^2+r_b^2+r_c^2 = \frac{4r^2(a^2+b^2+c^2)}{(a+b+c)^2}.$$
∎
This converts the geometric problem into a symmetric algebraic expression; skipping the expansion risks an incorrect coefficient.
Claim 4
The sum of the areas of the four circles equals
$$\frac{\pi(a^2+b^2+c^2)\bigl(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\bigr)} {(a+b+c)^4}.$$
Proof
The total area equals
$$\pi(r^2+r_a^2+r_b^2+r_c^2).$$
By Claim 3,
$$\pi(r^2+r_a^2+r_b^2+r_c^2) = \frac{4\pi r^2(a^2+b^2+c^2)}{(a+b+c)^2}.$$
Using
$$\Delta=rs,$$
and
$$s=\frac{a+b+c}{2},$$
we obtain
$$r=\frac{\Delta}{s} = \frac{2\Delta}{a+b+c}.$$
Hence
$$r^2=\frac{4\Delta^2}{(a+b+c)^2}.$$
Substituting,
$$\pi(r^2+r_a^2+r_b^2+r_c^2) = \frac{16\pi\Delta^2(a^2+b^2+c^2)}{(a+b+c)^4}.$$
Heron's formula gives
$$\Delta^2=s(s-a)(s-b)(s-c).$$
Expanding this standard symmetric form yields
$$16\Delta^2 = 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4.$$
Therefore
$$\pi(r^2+r_a^2+r_b^2+r_c^2) = \frac{ \pi(a^2+b^2+c^2) \bigl( 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4 \bigr) }{(a+b+c)^4}.$$
∎
This completes the computation; using Heron's formula without converting it into the symmetric polynomial form would not satisfy the requirement that the answer depend only on $a,b,c$.
Hence the sum of the areas of the four circles is
$$\boxed{ \frac{ \pi(a^2+b^2+c^2) \bigl( 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4 \bigr) }{(a+b+c)^4} }.$$
Verification of Key Steps
The first delicate step is the computation of the similarity ratio.
Independently recompute it. The altitude from $A$ to $BC$ equals
$$h_a=\frac{2\Delta}{a}.$$
The two parallel tangents to the incircle have separation $2r$, because the center lies midway between them and each tangent is distance $r$ from the center. Thus the smaller altitude is
$$h_a-2r.$$
The similarity ratio is therefore
$$\frac{h_a-2r}{h_a} = 1-\frac{2r}{h_a}.$$
Since
$$h_a=\frac{2rs}{a},$$
the ratio becomes
$$1-\frac{a}{s} = \frac{s-a}{s}.$$
A careless argument often replaces the separation by $r$ instead of $2r$, which would incorrectly produce
$$\frac{2s-a}{2s}.$$
The second delicate step is the algebraic simplification in Claim 3.
Compute directly:
$$\begin{aligned} (s-a)^2+(s-b)^2+(s-c)^2 &= 3s^2-2s(a+b+c)+(a^2+b^2+c^2) \ &= 3s^2-4s^2+(a^2+b^2+c^2) \ &= a^2+b^2+c^2-s^2. \end{aligned}$$
Then
$$1+\frac{a^2+b^2+c^2-s^2}{s^2} = \frac{a^2+b^2+c^2}{s^2}.$$
Dropping the initial $1$ or mishandling the sign of $s^2$ would produce an incorrect coefficient.
The third delicate step is the symmetric form of Heron's formula.
Starting from
$$16\Delta^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c),$$
group factors:
$$[(b+c)^2-a^2][a^2-(b-c)^2].$$
Expanding,
$$\begin{aligned} &\bigl(b^2+2bc+c^2-a^2\bigr) \bigl(a^2-b^2+2bc-c^2\bigr) \ &= 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4. \end{aligned}$$
A sign error in either factor reverses one quartic term and destroys symmetry.
Alternative Approaches
A coordinate approach is possible. Place
$$BC=a$$
on the $x$ axis and express the incenter using barycentric or Cartesian coordinates. The tangent parallel to $BC$ then has a simple equation, and the vertices of the smaller triangle can be obtained by intersecting this line with the sides $AB$ and $AC$. One may compute the side lengths of the smaller triangle explicitly and derive its inradius from the formula
$$r=\frac{\Delta}{s}.$$
This method is computationally longer and obscures the geometric reason for the appearance of the factors
$$\frac{s-a}{s},\quad \frac{s-b}{s},\quad \frac{s-c}{s}.$$
Another approach uses homothety centered at each vertex. The tangent parallel to $BC$ determines a homothety carrying $\triangle ABC$ to the corner triangle near $A$. The ratio can be obtained by comparing distances between corresponding parallel sides. This viewpoint is geometrically elegant, but it still requires the same computation of the distance between the two tangents, so it ultimately reduces to the same core idea as the main proof.