Good triangles in a circle

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Good triangles in a circle

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Source: ISL 2018 G3

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18 posts

#1 h Jul 17, 2019, 12:13 PM • 2 Y

Y by Adventure10, Mango247

A circle $\omega$ with radius $1$ is given. A collection $T$ of triangles is called good, if the following conditions hold:

each triangle from $T$ is inscribed in $\omega$ ;
no two triangles from $T$ have a common interior point.

Determine all positive real numbers $t$ such that, for each positive integer $n$ , there exists a good collection of $n$ triangles, each of perimeter greater than $t$ .

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test20

#2 h Jul 17, 2019, 3:57 PM • 1 Y

Y by Adventure10

A really nice problem. A kind of geometric inequality.

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Sepled

#3 h Jul 17, 2019, 4:09 PM • 5 Y

Y by Reef334, ArMath, Infinityfun, Adventure10, Mango247

The answer is $t \leq 4$ .

bound
construction

This post has been edited 1 time. Last edited by Sepled, Jul 17, 2019, 4:10 PM

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#4 h Jul 17, 2019, 4:33 PM • 4 Y

Y by v4913, starchan, Adventure10, MS_asdfgzxcvb

The answer is $0 < t \le 4$ .

First, we claim that all $t=4$ (and hence $t<4$ ) work. Indeed, fix $n$ and let $M > 2 n^3$ be a large integer. Construct a $2M$ -gon $A_0 A_1 \dots A_{2M-1}$ . Consider the triangles $\triangle A_0 A_1 A_M$ , $\triangle A_1 A_2 A_M$ , \dots, up to $\triangle A_{n-1} A_n A_M$ .

$[asy] size(6cm); draw(unitcircle); pair A = dir(170); pair B = dir(165); pair C = dir(0); dot("$A_{k-1}$", A, dir(180)); dot("$A_k$", B, dir(150)); dot("$A_M$", C, dir(C)); filldraw(A--B--C--cycle, rgb(.8,.8,.8), black); dot("$A_0$", dir(180), dir(225)); [/asy]$

The semi-perimeter of the $k$ th triangle ( $1 \le k \le n$ ) is $\begin{align*} s_k &= \sin\left( \frac{(M+(k-1))\pi}{2M} \right) + \sin\left( \frac{(M-k)\pi}{2M} \right) + \sin \left( \frac{\pi}{2M} \right) \\ &= \sin\left( \frac{\pi}{2M} \right) + \cos\left( \frac{(k-1)\pi}{2M} \right) + \cos\left( \frac{k\pi}{2M} \right) \\ &= \left( \frac{\pi}{2M} - O(M^{-3}) \right) + \left( 2 - O\left( (kM^{-1})^2 \right) \right) \\ &= 2 + \frac{\pi}{2M} - o(M^{-1}). \end{align*}$ where we have used Taylor's theorem in the form $\sin x = x - O(x^{-3})$ and $\cos x = 1 - O(x^{-2})$ . As $M \to \infty$ , we see $s_k > 2$ for all $1 \le k \le n$ , and so all the triangles have perimeter greater than $4$ .

Remark: Many other possible constructions exist for $t=4$ . Ankan gives the following one: initially start with a right triangle $ABC$ , with hypotenuse $BC$ , and $A$ near $B$ . The idea is to replace $ABC$ with two triangles $AXC$ and $XBC$ , with $X$ chosen on arc $\widehat{AB}$ , really close to $B$ , so that the perimeter of $AXC$ is greater than $4$ . Repeat.

On the other hand, suppose $t > 4+\varepsilon$ for some $\varepsilon > 0$ .

Claim: Any triangle with perimeter at least $t$ has area exceeding $\frac{1}{2} \varepsilon^{3/2}$ .

Proof. Let $a$ , $b$ , $c$ be the side lengths and $s$ the semiperimeter. We have $s \ge 2+\varepsilon/2$ and $a,b,c \le 2$ , hence $\min(s-a,s-b,s-c) > \varepsilon/2$ , so the area is at least $\sqrt{s(s-a)(s-b)(s-c)} > \sqrt{2 \cdot (\varepsilon/2)^3}$ by Heron's formula. $\blacksquare$

Thus there can be at most finitely many such triangles.

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#5 h Jul 17, 2019, 5:53 PM • 1 Y

Y by Adventure10

Answer $0<t\leq4$
t greater than 4 not possible as
Now it is enough to show that t=4 works

This post has been edited 1 time. Last edited by BOBTHEGR8, Jul 18, 2019, 8:15 AM

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Abderrahmane_Driouch

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Abderrahmane_Driouch

#6 h Jul 17, 2019, 5:59 PM • 2 Y

Y by Adventure10, Mango247

BOBTHEGR8 wrote:

Answer $0<t\leq4$
t greater than 4 not possible as
Now it is enough to show that t=4 works

We should have perimetet $\ge 4$

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#7 h Jul 18, 2019, 8:15 AM • 1 Y

Y by Adventure10

@ above ,its a typo

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#8 h Jan 6, 2023, 11:56 PM

Y by

The answer is all $t\in (0;4].$ Let $P(XYZ)$ denotes the perimeter of $\triangle XYZ.$

Construction. It's enough to consider case $t=4.$ Let $\omega '$ be a cemicircle of $\omega$ with diameter $AB_{n+1}.$ Pick $B_1\in \omega ';$ by the triangle inequality $P(AB_{n+1}B_1)>2\cdot |AB_{n+1}|=4.$ Next, for $i=2,3,\dots ,n$ because of $P(AB_{n+1}B_{i-1})> 4$ we can choose take $B_i$ on arc $B_{i-1}B$ of $\omega '$ (sufficently near to the $B$ ) for which $P(AB_{i-1}B_i)>4.$ Thus the collection of $\triangle AB_iB_{i+1}$ for $i=1,2,\dots ,n$ is good.

Upper bound. Now assume there is a good collection with $N$ triangles, each of perimeter greater than $4+\varepsilon .$ Pick an arbitrary triangle $ABC$ from collection; since each of it's side has length at most $2,$ the shortest side has length at least $\varepsilon.$ Using formula connecting area of triangle and it's circumradius we obtain $\text{area}(ABC)>\frac{\varepsilon^3}{4}.$ Since triangles of collection are disjoint, it follows that $N<\frac{4\pi}{\varepsilon^3}$ and hence the conclusion.

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awesomeming327.

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awesomeming327.

#9 h Mar 6, 2023, 5:06 AM

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The answer is $t\le 4$ . Note that for each triangle with lengths $a$ , $b$ , and $c$ we have $a,b,c\le 2$ so $a,b,c\ge t-4$ so if $t>4$ then note that $A=\frac{abc}{4R}=\frac{abc}{4}\ge \frac{(t-4)^3}{4}$ which means that enough triangles will cover the circle so there can't be infinite. If $t< 4$ then we first start by noting that any right triangle works, so let $AB$ be diameter and $C_1$ be any point on the circle and let it be close to $A$ . We will split this into $AC_2B$ and $C_2C_1B$ . Let $d$ be the perimeter minus $4$ . Then, let $AC_2=\frac{d}{2}$ and we get $C_1C_2\ge BC_1-\tfrac{d}2$ and $AC_2\ge AC-\tfrac{d}2$ so $AC_2+C_1C_2+AC_1\ge AC+AC_1+BC_1-d=4$ so we're done.

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dkedu

#10 h Jan 27, 2024, 9:31 PM

Y by

We claim the answer is $t \le 4$ .

If $t>4$ we have the area of a triangle is $\sqrt{s(s-a)(s-b)(s-c)} \ge \sqrt{\frac{t(t-4)^3)}{16}} > 0$ , so if $t >4$ we get that the triangles will have an area that is bounded below therefore if we pick $n$ large enough the sum of the areas of the triangles will be larger than the area of the circle and we will have a contradiction.

Now if $t \le 4$ , we just need to provide a construction for $t = 4$ . Let $\overline{AB}$ be a diameter. Pick $C$ on the circle close to $B$ . Note that the perimeter of this triangle must be greater than $4$ since $p = AC + BC + AB > 2AB = 4$ . Now we want to show that we can divide this triangle into two triangles with perimeter at least $4$ . To do this we simply pick $C'$ between $B$ and $C$ such that $BC' > \frac{p-4}{2}$ which obviously exists. Now realize that $CC' > BC - \frac{p-4}{2}$ and $AC' > AB - \frac{p-4}{2}$ so summing gives $AC + C'C + AC' > AC + BC - \frac{p-4}{2} + AB - \frac{p-4}{2} = p - (p-4) = 4$ . So we can infinitely split triangles of perimeter greater than $4$ so we are done.

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#11 h Jun 16, 2024, 3:51 PM

Y by

The answer is $\boxed{(0,4]}$
Firstly, we will prove that for all $t \leq 4$ such a collection exists.
Let $AB$ be the diameter of a circle. Delete the semicircle below $AB$
Now we will introduce a function $f: P \to P$ , where $P$ is a set of all points on the remaining semicircle
Let $f(X)$ be such a point $Y$ on arc $BX$ , that $BX+XY+YB=4$ (if there are multiple such points, choose the closest to $X$ ). Note that such a point doesn't exist for all $X$ , so when it doesn't exist, function is just not defined
Now note that our function is continious and $f(A)=A$ . Consider $g(X)=f^n(X)$ . It is also continious and also $g(A)=A$ . Then for any point $T$ close enough to $A$ there exists such a $M$ , that $f(M)=T$ . But then take triangles $BMf(M),Bf(M)f(f(M)),\ldots,Bf^{n-1}(M)f^n(M)$

For $t>4$ , note that by Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)} \geq \sqrt{p(p-2)(p-2)(p-2)} > \sqrt{\frac{t}{2}(\frac{t}{2}-2)^3}=C>0$ , and so not more than $\frac{\pi}{C}$ triangles can be in the collection

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Ritwin

#12 h Feb 28, 2025, 5:20 AM

Y by

Arbitrary-sized collections exist exactly when $t \leq 4$ . We prove this in two parts. Note the following fact.

Corollary [extended law of sines]: Suppose a triangle inscribed in $\omega$ has angles $\alpha$ , $\beta$ , and $\gamma$ . Then the triangle has perimeter $2 (\sin \alpha + \sin \beta + \sin \gamma)$ and area $2 \sin \alpha \sin \beta \sin \gamma$ .

$\boldsymbol{\color{blue} t > 4}$ is impossible. Let $t = 4 + \varepsilon$ . It suffices to prove for arbitrarily small $\varepsilon$ . We make the following claim.

Claim: Any triangle in a good collection must have every angle $\theta = \Omega(\varepsilon)$ .
Proof. Suppose a triangle inscribed in $\omega$ has smallest angle $\theta$ . Move the vertex $P$ with angle $\theta$ so that the triangle becomes isosceles with $P$ as its apex; this only increases its perimeter.
Now the triangle has angles $90^{\circ} - \tfrac\theta2$ twice and $\theta$ , so its perimeter is given by $2 \bigl(\sin \theta + 2 \cos \tfrac\theta2\bigr) = 4 + 2 \theta + O(\theta^2) > 4 + \varepsilon.$ This means $\theta > \tfrac12 \varepsilon - O(\varepsilon^2)$ , which we abbreviate as $\theta = \Omega(\varepsilon)$ . $\square$

On the other hand, this means every triangle in a good collection has area at least $2 (\sin \theta)^3 = \Omega(\varepsilon^3)$ . In particular, there is a positive lower bound $\kappa$ . Then every good collection must have at most $\pi/\kappa$ triangles, which is finite, contradiction. $\blacksquare$

$\boldsymbol{\color{blue} t \leq 4}$ is possible. It suffices to show $t = 4$ is possible. We provide a construction, describing the triangles by their angles. Let $\varepsilon > 0$ be a small number we will pick later. Consider the following:

For each $k \in \{1, 2, \ldots, n\}$ , add a triangle with angles $\varepsilon$ , $90^{\circ} + (k-1)\varepsilon$ , and $90^{\circ} - k \varepsilon$ .

In particular, this only adds triangles in half of the circle, which is okay. It's clear that as long as $k \varepsilon < 90^{\circ}$ , this procedure creates a good collection of triangles. It suffices to show that by picking $\varepsilon$ sufficiently small, every triangle will have perimeter greater than $4$ . But this is quick: by the lemma, triangle $k$ has perimeter $2 (\sin \varepsilon + \cos (k-1)\varepsilon + \cos k\varepsilon) = 4 + 2 \varepsilon - O_k(\varepsilon^2) = 4 + 2 \varepsilon - O_n(\varepsilon^2),$ so picking a sufficiently small $\varepsilon$ works. $\blacksquare$

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#13 h Mar 2, 2025, 5:01 AM

Y by

The answer is $t \leq 4$ .

Construction: For an $\varepsilon > 0$ yet to be determined, consider $n + 1$ points $A_0, A_1, \dots, A_n$ such that the arc $\widehat{A_iA_{i+1}} = 2\varepsilon$ for each $i$ , and let $P$ be the point symmetrically opposite on the circle. I claim that we can pick the triangles $PA_iA_{i+1}$ for each $0 \leq i \leq n-1$ by a suitable choice of $\varepsilon$ .

Clearly, the triangle with the smallest perimeter is the one on the edge, which has side lengths $2\sin \varepsilon$ , $2 \sin\left(90^\circ - \frac{n\varepsilon}2\right), 2\left(90^\circ - \frac{(n-2)\varepsilon}2\right)$ . So it suffices to show that there is a $\varepsilon > 0$ such that
$\sin \varepsilon + \cos\left(\frac{n\varepsilon}2\right) + \cos\left(\frac{(n-2)\varepsilon}2\right) > 2.$ Let $f(x) = \sin x + \cos\left(\frac{nx}2\right) + \cos\left(\frac{(n-2)x}2\right)$ and observe $f(0) = 2$ . Also, $f'(x) = \cos x - \frac nx \sin \left(\frac{nx}2\right) - \frac{n-2}2\sin\left(\frac{nx}2\right) \geq 0$ on some interval $[0, \varepsilon_0]$ . Thus at $\varepsilon = \varepsilon_0$ , $f(\varepsilon_0) > f(0) = 2$ , so this yields the result.

Construction: Assume that there is a $\varepsilon > 0$ such that there exists a construction with every triangle having perimeter greater than $4+2\varepsilon$ . Notice that because the $n$ triangles are disjoint, one has area at most $\frac{\pi}n$ .

Claim: There exists a side of a triangle in $T$ measuring less than $\varepsilon$ .

Proof: Otherwise, consider the area of the smallest triangle, which is given by $[\Delta] = \frac 12 S_1 S_2 \sin \theta > \frac 12\varepsilon^2 \cdot \frac{\varepsilon}{2R} = \frac 14 \varepsilon^3.$ Picking $\frac{\pi}n < \frac{\varepsilon^3}4$ yields a contradiction. $\blacksquare$

So this triangle $\Delta$ has perimeter less than $2+2+\varepsilon = 4+\varepsilon < 4+2\varepsilon$ , contradiction.

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