China Team Selection Test 2015 TST 1 Day 2 Q1

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sqing

41157 posts

sqing

#1 h Mar 14, 2015, 7:16 AM • 2 Y

Y by Adventure10, Mango247

Prove that : For each integer $n \ge 3$ , there exists the positive integers $a_1<a_2< \cdots <a_n$ , such that for $i=1,2,\cdots,n-2$ , With $a_{i},a_{i+1},a_{i+2}$ may be formed as a triangle side length , and the area of the triangle is a positive integer.

This post has been edited 1 time. Last edited by sqing, Mar 14, 2015, 7:17 AM

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mssmath

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mssmath

#2 h Mar 14, 2015, 7:15 PM • 2 Y

Y by Adventure10, Mango247

Does anybody happen to know the solution to this problem? In particular it doesn't appear to be easy for even n=4.

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MellowMelon

5850 posts

MellowMelon

#3 h Mar 14, 2015, 8:58 PM • 2 Y

Y by Adventure10, Mango247

The failure of SSA similarity is the key here. A sketch: Suppose we have an obtuse triangle $ABC$ with rational sides, obtuse angle at $C$ , and shortest side $BC$ . If we replace $C$ with its reflection over the altitude from $A$ , the new triangle has the same altitude and rational sides. You can determine how many times this construction can be iterated by looking at angles.

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Wolstenholme

543 posts

Wolstenholme

#4 h Mar 24, 2015, 9:51 PM • 9 Y

Y by vinhhop, langkhach11112, jt314, TheBeatlesVN, mijail, Kobayashi, hakN, Adventure10, Mango247

MellowMelon's post outlines the idea that solves the problem, but here's some more detail:

First replace the word "integer" in the problem with the word "rational" (we can do this by scaling each triangle).

Lemma 1 (MellowMelon): Suppose we have an obtuse triangle $ABC$ with rational sides, obtuse angle at $C$ , and shortest side $BC$ . If we replace $C$ with its reflection over the altitude from $A$ , the new triangle has the same altitude and rational sides.

Proof:

Let $c = AB, b = CA, a = BC,$ and let $d$ be the length of the altitude from $A.$ Also let $Z$ be the foot of the $A$ -altitude on $BC$ and let $x = CZ.$ It is clear that $d^2 + (a + x)^2 = c^2$ and $d^2 + x^2 = b^2$ so by subtracting these equations we find that $x = \frac{c^2 - a^2 - b^2}{2a} \in \mathbb{Q}$ as desired.

Now we return to the problem. Suppose we have an obtuse triangle with smallest angle $\alpha$ and second smallest angle $\beta.$ Clearly the obtuse angle has measure $180 - \alpha - \beta.$ Upon performing the operation described in Lemma 1 we find that the new triangle has an angle with measure $180 - \alpha - 2\beta > 180 - 2(\alpha + \beta).$ We can perform this operation so long as the triangle we have is obtuse (and it is clear that if after the operation we obtain a new, obtuse triangle, the "new" side length is the longest side of the triangle. Therefore by forcing $\alpha + \beta$ to be sufficiently small (which is equivalent to the original obtuse angle being sufficiently large), we can perform arbitrarily many operations.

So to solve the problem, it suffices to find a triangle with rational side lengths and rational area whose largest angle is arbitrarily close to $180$ degrees. But this isn't hard - consider the triangle with side lengths $8(m + 1)^2$ and $((2m + 1)^2 + 1)(2m + 3)$ and $((2m + 3)^2 + 1)(2m + 1).$ The altitude to the shortest side's length can be computed to be $2(2m + 1)(2m + 3).$ It is clear that as we let $m$ be a sufficiently large rational number, our triangle becomes "arbitrarily obtuse" so we are done.

This post has been edited 2 times. Last edited by Wolstenholme, Mar 24, 2015, 9:53 PM

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TheBeatlesVN

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TheBeatlesVN

#5 h Feb 13, 2017, 12:51 PM • 2 Y

Y by Adventure10, Mango247

Wolstenholme wrote:

MellowMelon's post outlines the idea that solves the problem, but here's some more detail:

First replace the word "integer" in the problem with the word "rational" (we can do this by scaling each triangle).

Lemma 1 (MellowMelon): Suppose we have an obtuse triangle $ABC$ with rational sides, obtuse angle at $C$ , and shortest side $BC$ . If we replace $C$ with its reflection over the altitude from $A$ , the new triangle has the same altitude and rational sides.

Proof:

Let $c = AB, b = CA, a = BC,$ and let $d$ be the length of the altitude from $A.$ Also let $Z$ be the foot of the $A$ -altitude on $BC$ and let $x = CZ.$ It is clear that $d^2 + (a + x)^2 = c^2$ and $d^2 + x^2 = b^2$ so by subtracting these equations we find that $x = \frac{c^2 - a^2 - b^2}{2a} \in \mathbb{Q}$ as desired.

Now we return to the problem. Suppose we have an obtuse triangle with smallest angle $\alpha$ and second smallest angle $\beta.$ Clearly the obtuse angle has measure $180 - \alpha - \beta.$ Upon performing the operation described in Lemma 1 we find that the new triangle has an angle with measure $180 - \alpha - 2\beta > 180 - 2(\alpha + \beta).$ We can perform this operation so long as the triangle we have is obtuse (and it is clear that if after the operation we obtain a new, obtuse triangle, the "new" side length is the longest side of the triangle. Therefore by forcing $\alpha + \beta$ to be sufficiently small (which is equivalent to the original obtuse angle being sufficiently large), we can perform arbitrarily many operations.

So to solve the problem, it suffices to find a triangle with rational side lengths and rational area whose largest angle is arbitrarily close to $180$ degrees. But this isn't hard - consider the triangle with side lengths $8(m + 1)^2$ and $((2m + 1)^2 + 1)(2m + 3)$ and $((2m + 3)^2 + 1)(2m + 1).$ The altitude to the shortest side's length can be computed to be $2(2m + 1)(2m + 3).$ It is clear that as we let $m$ be a sufficiently large rational number, our triangle becomes "arbitrarily obtuse" so we are done.

Could you tell me the way you choose three sides of the triangle ( $8(m + 1)^2$ and $((2m + 1)^2 + 1)(2m + 3)$ and $((2m + 3)^2 + 1)(2m + 1)$ ), please ?????

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supercarry

101 posts

supercarry

#6 h Apr 22, 2023, 7:30 AM

Y by

I think a reasonable way to find out the construction is to make the length of three side satisfying $x+y-z=2$ . To make the area is an integer, we may use Heron formula, which implies $(x-1)(y-1)(x+y-1)$ is square. An intuitive thought is to make $x-1=s^2, y-1=t^2$ . Hence $s^2+t^2+1$ is a square, and we can make $s=2n^2, t=2n$

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sttsmet

132 posts

sttsmet

#7 h Mar 19, 2025, 4:12 PM

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For the fixed $n$ , let $a, b$ the the last two terms in the sequence. We want to construct the n+1-th term. Skale everything by a factor of $k$ ( $k$ to be defined later).

I want to construct an obtuse triangle with sides $ka, kb, (d_1 + d_2 )$ satisfying the hypothesis. I just need to solve the system of equations:
$k^2a^2 = u^2 + d_{1}^2$ and $k^2b^2 = u^2 + d_{2}^2$ such that $d_1 + d_2 > a, b$ This is very easy :-D

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