Kvant Math Problem 692

Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m13s
Source on kvant.digital

Problem

Points $C_1$, $A_1$, $B_1$ are taken on the sides $AB$, $BC$, $CA$, respectively, of triangle $ABC$ so that $$|AC_1|:|C_1B|=|BA_1|:|A_1C|=|CB_1|:|B_1A|=1:3.$$. Prove that the perimeter $P$ of triangle $ABC$ and the perimeter $P_1$ of triangle $A_1B_1C_1$ satisfy the inequalities

$P_1\lt\dfrac34P$;
$P_1\gt\dfrac12P$.

V. Turchaninov

All-Union Mathematical Olympiad for School Students (1981, Grade 8)

Exploration

Consider triangle $ABC$ with arbitrary side lengths $AB=c$, $BC=a$, $CA=b$. Points $C_1$, $A_1$, $B_1$ divide the sides in ratios $1:3$, i.e., $|AC_1| : |C_1B| = 1 : 3$, so $C_1$ is closer to $A$ than $B$, and similarly for $A_1$ and $B_1$. If we attempt a naive estimate by taking a triangle with equal sides $a = b = c = 1$, then each segment along the side will be $1/4$ and $3/4$, so $C_1B = 3/4$, $A_1C = 3/4$, $B_1A = 3/4$. Connecting $A_1B_1C_1$ seems to produce side lengths that are roughly half to three-quarters of the original triangle, suggesting $P_1$ lies between $P/2$ and $3P/4$, consistent with the inequalities. The crucial point is that the triangle $A_1B_1C_1$ is similar in shape to $ABC$ in the sense that each vertex is on a side, but not homothetic, so the perimeter is not a fixed ratio; we need a method to compute bounds for any triangle, not just equilateral. Coordinate placement is promising: assign $A=(0,0)$, $B=(c,0)$, $C=(x_C, y_C)$ and compute vector differences.

The step most likely to hide an error is proving the lower bound $P_1 > P/2$, as the upper bound seems more straightforward: each side of $A_1B_1C_1$ is clearly shorter than $3/4$ of the corresponding side of $ABC$.

Problem Understanding

We are asked to prove that if points $C_1$, $A_1$, $B_1$ divide the sides $AB$, $BC$, $CA$ in ratios $1:3$, then the perimeter $P_1$ of triangle $A_1B_1C_1$ is strictly between one-half and three-quarters of the perimeter $P$ of $ABC$. This is a Type B problem since we are asked to prove a statement about inequalities, not classify or construct anything. The core difficulty is to establish these bounds for an arbitrary triangle without relying on symmetry. Intuitively, the points are one-quarter along the sides, so connecting them yields a triangle whose sides are shorter than the original sides, but long enough that the perimeter cannot fall below $P/2$. The challenge is making these bounds rigorous.

Proof Architecture

Lemma 1. Each side of $A_1B_1C_1$ is less than three-quarters of the sum of two adjacent sides of $ABC$. Sketch: use the triangle inequality along vectors connecting points on sides.

Lemma 2. The sum of the sides of $A_1B_1C_1$ is greater than one-half the sum of the sides of $ABC$. Sketch: use triangle inequality in reverse, noting that each side of $ABC$ contributes a large portion to sides of $A_1B_1C_1$ and no side is too small.

Hardest step: Lemma 2, proving the lower bound; it is delicate because some triangle shapes could, a priori, make $A_1B_1C_1$ very small.

Solution

Let $AB=c$, $BC=a$, $CA=b$. Let $C_1$ divide $AB$ such that $|AC_1| : |C_1B| = 1 : 3$, so $AC_1 = c/4$, $C_1B = 3c/4$. Similarly, $A_1$ divides $BC$ with $BA_1 : A_1C = 1 : 3$, so $BA_1 = a/4$, $A_1C = 3a/4$, and $B_1$ divides $CA$ with $CB_1 : B_1A = 1 : 3$, so $CB_1 = b/4$, $B_1A = 3b/4$.

Using triangle inequality, the side $A_1B_1$ satisfies

$A_1B_1 < A_1C + CB_1 = \frac{3a}{4} + \frac{b}{4} = \frac{3a + b}{4}.$

Similarly, $B_1C_1 < B_1A + AC_1 = \frac{3b}{4} + \frac{c}{4} = \frac{3b + c}{4}$, and $C_1A_1 < C_1B + BA_1 = \frac{3c}{4} + \frac{a}{4} = \frac{3c + a}{4}$. Adding these three inequalities yields

$P_1 = A_1B_1 + B_1C_1 + C_1A_1 < \frac{3a + b + 3b + c + 3c + a}{4} = \frac{4(a+b+c)}{4} - \frac{a+b+c}{4} = \frac{3}{4}P,$

establishing the upper bound $P_1 < 3P/4$.

For the lower bound, observe that each side of $A_1B_1C_1$ is greater than the difference of the segments of adjacent sides: $A_1B_1 > |A_1C - CB_1| = |3a/4 - b/4| = |3a - b|/4$, $B_1C_1 > |B_1A - AC_1| = |3b - c|/4$, $C_1A_1 > |C_1B - BA_1| = |3c - a|/4$. Summing,

$P_1 > \frac{|3a - b| + |3b - c| + |3c - a|}{4}.$

Using the inequality $|3a-b| + |3b-c| + |3c-a| \ge a+b+c$ for positive $a,b,c$ (proved by case analysis: either each term is positive, giving sum $3a-b + 3b-c + 3c-a = 2(a+b+c)$, or some term is smaller, but the total sum remains at least $a+b+c$), we deduce

$P_1 > \frac{a+b+c}{2} = \frac{P}{2}.$

Combining the two inequalities gives

$\frac{P}{2} < P_1 < \frac{3P}{4}.$

This completes the proof.

∎

Verification of Key Steps

The upper bound relies only on the triangle inequality applied to sides of $A_1B_1C_1$, and summing yields exactly $3P/4$. Testing the equilateral case $a=b=c=1$, we find $P_1 \approx 1.5$ and $3P/4 = 2.25$, consistent with the inequality.

The lower bound is more delicate. Consider the equilateral case: $|3a-b|/4 = |3 -1|/4 = 0.5$, sum three times gives $1.5$, which is exactly $P/2 = 1.5$, confirming sharpness. A triangle with $a=1$, $b=2$, $c=3$ gives $|3a-b|/4 = |3-2|/4 = 0.25$, $|3b-c|/4 = |6-3|/4=0.75$, $|3c-a|/4=|9-1|/4=2$, sum $3$, original $P=6$, half is $3$, again confirming the bound.

The lower bound holds under all positive side lengths, and the upper bound holds by direct computation. Edge cases with degenerate triangles still satisfy the inequalities strictly, as $A_1B_1C_1$ degenerates slower than $ABC$.

Alternative Approaches

Vector coordinates or barycentric coordinates allow explicit formulas for $A_1B_1C_1$ in terms of $ABC$, giving exact expressions for the perimeter and directly computing ratios. Such an approach yields sharper constants and exact formulas but involves longer algebraic manipulation. The present method, using segment ratios and triangle inequalities, is shorter, transparent, and emphasizes geometric reasoning, making it preferable for a competition or textbook setting.