Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

Divide the segment $A B$ into two parts with the internal point $C$ and construct a square $A C D E$ on $A C$ , and an equilateral triangle $C B F$ on $C B$ on the same side of the line $A B$ . It is to be proven that the area of the pentagon $A B F D E$ is always greater than one third of the area of the square that can be constructed over $A B$ .

Let's translate the problem and the solution step by step:

1. **Divide the segment $A B$ into two parts with the internal point $C$ **:
- Let $A C = x$ and $C B = y$ . Therefore, $A B = x + y$ .

2. **Construct a square $A C D E$ on $A C$ **:
- The side length of the square $A C D E$ is $x$ .
- The area of the square $A C D E$ is $x^2$ .

3. **Construct an equilateral triangle $C B F$ on $C B$ **:
- The side length of the equilateral triangle $C B F$ is $y$ .
- The area of the equilateral triangle $C B F$ is $\frac{\sqrt{3}}{4} y^2$ .

4. **Calculate the area of the pentagon $A B F D E$ **:
- The pentagon $A B F D E$ consists of the square $A C D E$ and the equilateral triangle $C B F$ .
- The area of the pentagon $A B F D E$ is $x^2 + \frac{\sqrt{3}}{4} y^2$ .

5. **Calculate the area of the square that can be constructed over $A B$ **:
- The side length of the square over $A B$ is $x + y$ .
- The area of the square over $A B$ is $(x + y)^2$ .
- One third of the area of the square over $A B$ is $\frac{1}{3} (x + y)^2$ .

6. **Prove that the area of the pentagon $A B F D E$ is always greater than one third of the area of the square over $A B$ **:
- We need to show that $x^2 + \frac{\sqrt{3}}{4} y^2 > \frac{1}{3} (x + y)^2$ .
- Expand the right-hand side: $\frac{1}{3} (x + y)^2 = \frac{1}{3} (x^2 + 2xy + y^2) = \frac{1}{3} x^2 + \frac{2}{3} xy + \frac{1}{3} y^2$ .
- We need to show that $x^2 + \frac{\sqrt{3}}{4} y^2 > \frac{1}{3} x^2 + \frac{2}{3} xy + \frac{1}{3} y^2$ .
- Rearrange the inequality: $x^2 + \frac{\sqrt{3}}{4} y^2 - \frac{1}{3} x^2 - \frac{2}{3} xy - \frac{1}{3} y^2 > 0$ .
- Simplify the left-hand side: $\left(1 - \frac{1}{3}\right) x^2 + \left(\frac{\sqrt{3}}{4} - \frac{1}{3}\right) y^2 - \frac{2}{3} xy > 0$ .
- This simplifies to: $\frac{2}{3} x^2 + \left(\frac{\sqrt{3}}{4} - \frac{1}{3}\right) y^2 - \frac{2}{3} xy > 0$ .
- Since $\frac{\sqrt{3}}{4} \approx 0.433$ and $\frac{1}{3} \approx 0.333$ , we have $\frac{\sqrt{3}}{4} - \frac{1}{3} > 0$ .
- Therefore, the inequality holds for all positive values of $x$ and $y$ .

Thus, the area of the pentagon $A B F D E$ is always greater than one third of the area of the square that can be constructed over $A B$ .

Solution

I. solution. Let $AC = a, CB = b$ . Then $AB = a + b$ , and the perpendicular height of triangle $CDF$ from $CD = a$ is $b/2$ . Then - denoting the areas of the figures by the figures themselves - the following inequality must be proven:

$ABFD E = ACDE + CDF + CBF = a^2 + \frac{ab}{4} + \frac{b^2 \sqrt{3}}{4} > \frac{(a+b)^2}{3}$

in other words, that the difference between the left and right sides is positive:

$\frac{2a^2}{3} - \frac{5ab}{12} + \left(\frac{\sqrt{3}}{4} - \frac{1}{3}\right) b^2 > 0$

From the first two terms, completing the square:

$\frac{2a^2}{3} - \frac{5ab}{12} = \frac{2}{3}\left(a^2 - \frac{5}{8}ab\right) = \frac{2}{3}\left(a - \frac{5}{16}b\right)^2 - \frac{25}{384}b^2$

so the difference is

$\frac{2}{3}\left(a - \frac{5}{16}b\right)^2 + \left(\frac{\sqrt{3}}{4} - \frac{51}{128}\right) b^2$

Here, the coefficient of $b^2$ is positive, because

$\frac{\sqrt{3}}{4} = \frac{32\sqrt{3}}{128} = \frac{\sqrt{3072}}{128} > \frac{55}{128} > \frac{51}{128}$

so the expression is indeed always positive. This completes the proof.

László Márki (Budapest, Fazekas M. High School)

II. solution. Let $AB = c, AC = x$ , so $CB = c - x$ and according to the first solution, the area of the pentagon is:

$y = x^2 + \frac{x(c-x)}{4} + \frac{(c-x)^2 \sqrt{3}}{4} = \frac{3+\sqrt{3}}{4} x^2 - \frac{2\sqrt{3}-1}{4} cx + \frac{c^2 \sqrt{3}}{4}$

We seek the minimum value of $y$ , with the constraint that $x$ varies only from 0 to $c$ . If we find this to be greater than $c^2 / 3$ , then the statement of the problem is true.

For simpler calculations, let's denote the coefficients of $x^2, cx, c^2$ temporarily as $p, q, r$ respectively, so $y = px^2 + qcx + rc^2$ . By factoring and completing the square:

$\begin{gathered} y = p\left(x^2 + \frac{q}{p} cx + \frac{r}{p} c^2\right) = p\left[\left(x + \frac{qc}{2p}\right)^2 - \frac{q^2 c^2}{4p^2} + \frac{r}{p} c^2\right] = \\ = p\left(x + \frac{qc}{2p}\right)^2 + \frac{4pr - q^2}{4p} c^2 \end{gathered}$

and with the coefficients, the calculations yield:

$y = \frac{3+\sqrt{3}}{4}\left(x - \frac{7\sqrt{3}-9}{12} c\right)^2 + \frac{49\sqrt{3}-51}{96} c^2$

The minimum value of $y$ is obtained when the first term, which contains the variable alone, is 0. This happens when

$x = \frac{7\sqrt{3}-9}{12} c \quad (\approx 0.26c)$

which is within the allowed range for $x$ . In this case, $y$ is equal to the second term:

$y_{\min} = \frac{49\sqrt{3}-51}{96} c^2 = \frac{\sqrt{7203}-51}{96} c^2 > \frac{84-51}{96} c^2 = \frac{33}{96} c^2 > \frac{1}{3} c^2$

This completes the proof of the statement.

János Csirik (Orosháza, Táncsics M. High School)

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.