Gheorghe Țițeica 2025 Grade 7 P3

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Source: Gheorghe Țițeica 2025

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#1 h Mar 28, 2025, 8:41 PM

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Out of all the nondegenerate triangles with positive integer sides and perimeter $100$ , find the one with the smallest area.

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#2 h Mar 29, 2025, 3:19 PM

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If degenerate triangles were allowed, the optimal choice would be a collinear triangle. For a non-degenerate triangle we want to come as close as possible to collinearity; triangles with sidelengths 25, 26, 49 or 2, 49, 49 come to mind. Let us try to produce a rigorous argument for the claim that the latter sidelengths 2, 49, 49 do indeed produce the smallest possible area.

We compare the areas of a triangle with sidelengths $a, b,c$ and a triangle with sidelengths $a, b-1, c+1$ , assuming that $a \le b \le c$ and $a + b+ c = 100$ . Using Heron's formula, we see that it suffices to compare the terms
$50 \cdot (50-a)(50-b)(50-c)$ for the first triangle and
$50 \cdot (50-a)(50-(b-1))(50-(c+1))$ for the second triangle. As the first two factors are the same in both terms, we can reduce the comparison to the terms
$(50-b)(50-c)$ and
$(50-(b-1))(50-(c+1)).$ Subtracting the last term from the one before it, we get $c - b +1$ , which is positive due to $c \ge b$ . This shows that the area of the second triangle is smaller.

Now, if we have a non-degenerate triangle with sidelengths $a,b,c$ and $a \le b \le c$ and $a + b+ c=100$ and with minimal area, the above step from sidelengths $a,b,c$ to sidelengths $a, b-1, c+1$ must be barred for some reason (otherwise the area would not be minimal). The only possible reason for that is that a non-degenerate triangle with sidelengths $a, b-1, c+1$ does not exist due to a violation of the triangle inequalities. As the triangle inequalities for the $a$ - $b$ - $c$ -triangle are fulfilled, the only triangle inequality which can be violated for the hypothetical triangle with the new sidelengths is $a + (b-1) > c+1$ , meaning that in fact we have
$a + (b-1) \le c+1.$ As the triangle inequalities for the $a$ - $b$ - $c$ -triangle are fulfilled (due to its existence), we have $a+b>c$ and therefore
$a+b -1 \ge c.$ So, taking these inequalities together, we have
$c \le a+b-1 \le c+1$ and, adding 1, we get
$c+1 \le a+b \le c+2.$ Consequently, $a+b$ is equal either to $c+1$ or to $c+2$ . Assuming the first case, $a+b=c+1$ , we get
$100 = a+b+c = 2c +1$ which is not possible for an integer value of $c$ . So the second case must be true: $a+b=c+2$ . This leads us to
$100 = a+b+c = 2c + 2$ which means
$c=49$ and $a+b=51.$ We are almost done. We still need to find out what $a$ and $b$ are. As we already know that $c=49$ , the Heronian formula tells us that the area of our minimal triangle only depends on the term
$(50-a)(50-b)$ which is equivalent to $2500 - 50 (a+b) + ab.$ As $a+b$ is already fixed ( $a+b=51$ ), the only variable here is the product $ab$ . Now
$ab=a(51-a)=-(a-\tfrac{51}{2})^2+\tfrac{51^2}{4}.$ For the positive integer value of $a$ , we need to get as far away from $\tfrac{51}{2}$ as possible to minimalize $ab$ . As $a \le b$ , we must investigate the lower end of the spectrum for $a$ . The value $a = 1$ is forbidden, as $b$ would be $50$ then, violating $b \le c$ and the triangle inequality $a+c > b$ . The best acceptable choice is obviously $a=2$ . So we finally have
$a = 2 \quad \text{and} \quad {b=49} \quad \text{and} \quad {c=49}.$

This post has been edited 1 time. Last edited by Rainbow1971, Mar 29, 2025, 3:33 PM
Reason: Correction of minor error in the first paragraph

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