Kvant Math Problem 1007
The equality
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m12s
Source on kvant.digital
Problem
Prove that triangles with side lengths $a$, $b$, $c$ and $a_1$, $b_1$, $c_1$ are similar if and only if $$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a_1+b_1+c_1)(a+b+c)}.$$
V. P. Chichin
Exploration
The equality
$$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1} = \sqrt{(a+b+c)(a_1+b_1+c_1)} \tag{1}$$
strongly resembles the equality case of the Cauchy-Schwarz inequality. Indeed,
$$(\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2 \le (a+b+c)(a_1+b_1+c_1),$$
with equality exactly when
$$\frac{\sqrt a}{\sqrt{a_1}} = \frac{\sqrt b}{\sqrt{b_1}} = \frac{\sqrt c}{\sqrt{c_1}},$$
that is,
$$a=\lambda a_1,\qquad b=\lambda b_1,\qquad c=\lambda c_1.$$
If this argument is correct, the condition immediately forces the three corresponding sides to be proportional, which is precisely the criterion for similarity.
Before accepting this, it is worth checking that no triangle-specific subtlety is hidden. Take two non-similar triangles with sides $(3,4,5)$ and $(2,3,4)$. Then
$$\sqrt6+\sqrt{12}+\sqrt{20}\approx10.382,$$
while
$$\sqrt{(3+4+5)(2+3+4)} = \sqrt{108} \approx10.392.$$
The inequality is strict. For similar triangles, say $(3,4,5)$ and $(6,8,10)$,
$$\sqrt{18}+\sqrt{32}+\sqrt{50} = 3\sqrt2+4\sqrt2+5\sqrt2 = 12\sqrt2,$$
and
$$\sqrt{12\cdot24} = 12\sqrt2.$$
Equality holds.
The only potentially dangerous step is the characterization of equality in Cauchy-Schwarz. That must be proved carefully, because the entire argument rests on it.
Problem Understanding
We are given two triangles with side lengths $a,b,c$ and $a_1,b_1,c_1$. We must prove that the triangles are similar if and only if
$$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1} = \sqrt{(a+b+c)(a_1+b_1+c_1)}.$$
This is a Type B problem. The statement to be proved is already specified.
The core difficulty is to recognize that the given equality is exactly the equality case of the Cauchy-Schwarz inequality and then connect the resulting proportionality of side lengths with the criterion for similarity of triangles.
Proof Architecture
Consider the vectors
$$(\sqrt a,\sqrt b,\sqrt c), \qquad (\sqrt{a_1},\sqrt{b_1},\sqrt{c_1}).$$
Applying Cauchy-Schwarz to these vectors yields
$$(\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2 \le (a+b+c)(a_1+b_1+c_1).$$
Equality in Cauchy-Schwarz holds if and only if the two vectors are proportional; this gives
$$a=\lambda a_1,\quad b=\lambda b_1,\quad c=\lambda c_1.$$
Proportional corresponding side lengths are equivalent to similarity of triangles by the SSS similarity criterion.
The harder direction is proving that the given equality implies similarity. The step most likely to fail under scrutiny is the equality condition in Cauchy-Schwarz.
Solution
Suppose first that
$$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1} = \sqrt{(a+b+c)(a_1+b_1+c_1)}. \tag{1}$$
Apply the Cauchy-Schwarz inequality to the vectors
$$(\sqrt a,\sqrt b,\sqrt c) \quad\text{and}\quad (\sqrt{a_1},\sqrt{b_1},\sqrt{c_1}).$$
This gives
$$(\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2 \le (a+b+c)(a_1+b_1+c_1). \tag{2}$$
Since (1) states that equality holds in (2), we are in the equality case of the Cauchy-Schwarz inequality. Hence there exists a positive number $t$ such that
$$\sqrt a=t\sqrt{a_1},\qquad \sqrt b=t\sqrt{b_1},\qquad \sqrt c=t\sqrt{c_1}.$$
Squaring these relations yields
$$a=t^2a_1,\qquad b=t^2b_1,\qquad c=t^2c_1. \tag{3}$$
Thus the corresponding sides of the two triangles are proportional. By the SSS similarity criterion, the triangles are similar.
Conversely, suppose the triangles are similar. Then there exists a positive number $\lambda$ such that
$$a=\lambda a_1,\qquad b=\lambda b_1,\qquad c=\lambda c_1. \tag{4}$$
Using (4),
$$\begin{aligned} \sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1} &= \sqrt\lambda,a_1+\sqrt\lambda,b_1+\sqrt\lambda,c_1 \ &= \sqrt\lambda,(a_1+b_1+c_1). \end{aligned}$$
Also,
$$\begin{aligned} \sqrt{(a+b+c)(a_1+b_1+c_1)} &= \sqrt{\lambda(a_1+b_1+c_1)(a_1+b_1+c_1)} \ &= \sqrt\lambda,(a_1+b_1+c_1). \end{aligned}$$
The two expressions are equal, so
$$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1} = \sqrt{(a+b+c)(a_1+b_1+c_1)}.$$
Both implications have been proved. This completes the proof.
∎
Verification of Key Steps
The crucial step is the equality case of Cauchy-Schwarz. For vectors
$$x=(\sqrt a,\sqrt b,\sqrt c),\qquad y=(\sqrt{a_1},\sqrt{b_1},\sqrt{c_1}),$$
Cauchy-Schwarz states
$$(x\cdot y)^2\le (x\cdot x)(y\cdot y).$$
Equality holds precisely when $x$ and $y$ are linearly dependent. Since all coordinates are positive, there exists a positive number $t$ with $x=t y$. This gives
$$\sqrt a=t\sqrt{a_1},\quad \sqrt b=t\sqrt{b_1},\quad \sqrt c=t\sqrt{c_1},$$
and hence (3).
Another delicate point is the passage from (3) to similarity. Equation (3) gives
$$\frac a{a_1} = \frac b{b_1} = \frac c{c_1} = t^2.$$
The three corresponding sides are proportional. The SSS similarity criterion applies directly, so the triangles are similar. No angle information is required.
Finally, in the converse direction, substituting
$$a=\lambda a_1,\quad b=\lambda b_1,\quad c=\lambda c_1$$
must be carried out explicitly. Both sides reduce to
$$\sqrt\lambda,(a_1+b_1+c_1),$$
which confirms the equality without invoking any further theorem.
Alternative Approaches
Instead of citing the equality case of Cauchy-Schwarz directly, one may expand the difference between the two sides squared:
$$\begin{aligned} &(a+b+c)(a_1+b_1+c_1) -(\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2 \ &= (\sqrt{ab_1}-\sqrt{a_1b})^2 + (\sqrt{ac_1}-\sqrt{a_1c})^2 + (\sqrt{bc_1}-\sqrt{b_1c})^2. \end{aligned}$$
The given equality forces this sum of three nonnegative squares to vanish. Hence
$$\sqrt{ab_1}=\sqrt{a_1b},\qquad \sqrt{ac_1}=\sqrt{a_1c},\qquad \sqrt{bc_1}=\sqrt{b_1c},$$
which implies
$$\frac a{a_1} = \frac b{b_1} = \frac c{c_1}.$$
Similarity then follows from SSS.
The main proof is preferable because it identifies the condition immediately as the equality case of a standard inequality and reaches the conclusion in a single conceptual step.