Comparing inradii of triangles

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Comparing inradii of triangles

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Source: India TST 2016 Day 4 Problem 1

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#1 h Jul 22, 2016, 7:43 AM • 1 Y

Y by Adventure10

Let $ABC$ be an acute triangle with circumcircle $\Gamma$ . Let $A_1,B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC,CBA$ and $ACB$ of $\Gamma$ . Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$ .

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#2 h Jul 22, 2016, 8:02 AM • 2 Y

Y by biomathematics, Adventure10

wlogA>B>C, the problem is equivalent to sin(A/2)sin(B/2)sin(C/2)<=sin((B+C)/4)sin((C+A)/4)sin((A+B)/4),use the Karamata Inequality with lnsin(x),done

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#3 h Jul 23, 2016, 6:24 AM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Anant Mudgal (IND 2) gave a synthetic solution to this problem, for which he won a special prize.

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#4 h Jul 23, 2016, 8:22 AM • 11 Y

Y by Ankoganit, gavrilos, TheOneYouWant, biomathematics, amar_04, Euler1728, aops29, guptaamitu1, Adventure10, Mango247, ehuseyinyigit

Ok, the special prize might have been given because no one even tried synthetic; and gave an algebraic proof mostly.

We first recall the Euler Poncelet Formula, $d=\sqrt{R^2-2Rr}$ where the notations are what they usually mean. Note also that the incenter of $ABC$ is the DeLongChamps point of triangle $A_1B_1C_1$ . We also see that triangle $A_1B_1C_1$ is acute angled, being similar to the intouch triangle of $ABC$ . Now the result which we seek to prove is the following: In acute angled triangle $XYZ$ with orthocenter $H$ , incenter $I$ and circumcenter $O$ , we have $OH \ge OI$ .

The proof is as follows: Assume wlog that $\angle X$ is the median of the set $\{\angle X,\angle Y,\angle Z\}$ . Since $H,O$ are isogonal conjugates, we see that $YI$ bisects angle $OYH$ and $ZI$ bisects angle $OZH$ . Let $YH \cap ZO=P$ and $YO \cap ZH=Q$ . We claim that $HPOQ$ is a convex quadrilateral and $I$ lies in its interior. We can see that $YHOZ$ is a convex quadrilateral by our angle condition on $X$ . Therefore, $P,Q$ lies on opposite sides of $HO$ and hence $HPOQ$ is convex. Now, $I$ lies in its interior immediately follows. Observe that $\angle OPH=180^{\circ}-\angle Y >90^{\circ}$ and $\angle OQH=180^{\circ}-\angle Z>90^{\circ}$ . Let $OI$ extended meet the boundary of the quadrilateral $OPHQ$ at point $T$ . We clearly have $\angle OTH>\min\{\angle OPH,\angle OQH\}>90^{\circ}$ . Therefore, $OI \le OT \le OH$ . The result holds.

Comment This was the only time in India TSTs when a solution graded $9/10$ (due to some technical issue) got a special prize.

This post has been edited 3 times. Last edited by anantmudgal09, Jul 23, 2016, 10:11 AM

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#6 h Feb 15, 2017, 1:06 PM • 2 Y

Y by Adventure10, Mango247

This might be easy but somebody please give a complete trigonometric solution.

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#8 h Mar 11, 2017, 2:10 PM • 1 Y

Y by Adventure10

PRO2000 wrote:

This might be easy but somebody please give a complete trigonometric solution.

I think the inequality

$sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2}) \leq sin(\frac{A+B}{4})sin(\frac{A+C}{4})sin(\frac{C+B}{4})$

Can be reached easily by A.M-G.M. on $sin(\frac{A}{4})cos(\frac{B}{4})$ and $sin(\frac{B}{4})cos(\frac{A}{4})$

Which gives

$sin(\frac{A+B}{4}) \geq 2(sin(\frac{A}{4})cos(\frac{B}{4})sin(\frac{B}{4})cos(\frac{A}{4}))^\frac{1}{2}$

Now get the other two similar inequalities and multiply them to get the answer.

This post has been edited 1 time. Last edited by united_jay99, Mar 11, 2017, 2:10 PM

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#9 h Apr 2, 2018, 9:08 PM • 4 Y

Y by GoldGirl, Adventure10, Mango247, math_comb01

Solution:

We know that $\sum_\text{cyc} \cos A = 1 + \frac{r}{R}$ , so we just need that $\sum_\text{cyc} \cos A \le \sum_\text{cyc} \sin \frac A2$ .
But this is easy because $\sum_\text{cyc} \cos A = \sum_\text{cyc} \frac{\cos B + \cos C}{2} = \sum_\text{cyc} \cos \left( \frac{B-C}{2} \right) \sin \frac A2 \le \sum_\text{cyc} \sin \frac A2$

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#10 h Apr 2, 2018, 9:49 PM • 2 Y

Y by Adventure10, Mango247

From the IMOSL 1988 problem here (and my solution using Schur's

), we know that the area of $A_1 B_1 C_1$ is greater than or equal to that of $ABC$ . It's quite easy to see that the semi-perimeters of the two triangles is the same. Since $r = \frac{\Delta}{s}$ , the conclusion follows immediately.

EDIT: As @below highlights, this is not valid. Basically what I was assuming is if an arc doubles in length, so does its associated chord. This is, of course, false

This post has been edited 1 time. Last edited by Arthur., Apr 17, 2018, 6:16 PM

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#11 h Apr 2, 2018, 11:12 PM • 3 Y

Y by GoldGirl, Adventure10, Mango247

Arthur. wrote:

It's quite easy to see that the semi-perimeters of the two triangles is the same.

I don't think so.

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#12 h Apr 3, 2018, 12:57 AM • 2 Y

Y by GoldGirl, Adventure10

anantmudgal09 wrote:

Ok, the special prize might have been given because no one even tried synthetic; and gave an algebraic proof mostly.

We first recall the Euler Poncelet Formula, $d=\sqrt{R^2-2Rr}$ where the notations are what they usually mean. Note also that the incenter of $ABC$ is the DeLongChamps point of triangle $A_1B_1C_1$ . We also see that triangle $A_1B_1C_1$ is acute angled, being similar to the intouch triangle of $ABC$ . Now the result which we seek to prove is the following: In acute angled triangle $XYZ$ with orthocenter $H$ , incenter $I$ and circumcenter $O$ , we have $OH \ge OI$ .

The proof is as follows: Assume wlog that $\angle X$ is the median of the set $\{\angle X,\angle Y,\angle Z\}$ . Since $H,O$ are isogonal conjugates, we see that $YI$ bisects angle $OYH$ and $ZI$ bisects angle $OZH$ . Let $YH \cap ZO=P$ and $YO \cap ZH=Q$ . We claim that $HPOQ$ is a convex quadrilateral and $I$ lies in its interior. We can see that $YHOZ$ is a convex quadrilateral by our angle condition on $X$ . Therefore, $P,Q$ lies on opposite sides of $HO$ and hence $HPOQ$ is convex. Now, $I$ lies in its interior immediately follows. Observe that $\angle OPH=180^{\circ}-\angle Y >90^{\circ}$ and $\angle OQH=180^{\circ}-\angle Z>90^{\circ}$ . Let $OI$ extended meet the boundary of the quadrilateral $OPHQ$ at point $T$ . We clearly have $\angle OTH>\min\{\angle OPH,\angle OQH\}>90^{\circ}$ . Therefore, $OI \le OT \le OH$ . The result holds.

Comment This was the only time in India TSTs when a solution graded $9/10$ (due to some technical issue) got a special prize.

What is the name of the prize????

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#13 h Jan 18, 2025, 4:57 PM

Y by

Ehh,complex bashing can be used,as mentioned in egmo.

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#14 h Mar 26, 2025, 12:21 PM • 1 Y

Y by L13832

Very standard using Jensen's.

A trivial angle chase reveals that the angles of $\Delta A_{1}B_{1}C_{1}$ are $\left(\frac{B+C}{2}\right),\left(\frac{C+A}{2}\right),\left(\frac{A+B}{2}\right)$ . It is well known that $\boxed{r=4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)}$ in $\Delta ABC$ . Thus,
$r=4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)$ $r_{1}=4R\sin\left(\frac{A+B}{4}\right)\sin\left(\frac{B+C}{4}\right)\sin\left(\frac{C+A}{4}\right)$ where $r$ and $r_1$ are inradii of $\Delta ABC$ and $\Delta A_{1}B_{1}C_{1}$ respectively.
We wish to show that $r_{1} \geq r$ . Thus,
$r_{1} \geq r$ $\iff \boxed{\sin\left(\frac{A+B}{4}\right)\sin\left(\frac{B+C}{4}\right)\sin\left(\frac{C+A}{4}\right) \geq \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)}$ Claim: $\color{blue}{\sin\left(\frac{A+B}{4}\right)\sin\left(\frac{B+C}{4}\right)\sin\left(\frac{C+A}{4}\right) \geq \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)}$
Proof: Note that $\sin x$ is concave in the interval $[0,\pi]$ . Thus by Jensen's Inequality we have
$\sin\left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right) \leq 2\sin\left(\frac{A+B}{4}\right) \cdots (1)$ $\sin\left(\frac{B}{2}\right)+\sin\left(\frac{C}{2}\right) \leq 2\sin\left(\frac{B+C}{4}\right) \cdots (2)$ $\sin\left(\frac{C}{2}\right)+\sin\left(\frac{A}{2}\right) \leq 2\sin\left(\frac{C+A}{4}\right) \cdots (3)$ Multiplying all, we get
$8\sin\left(\frac{A+B}{4}\right)\sin\left(\frac{B+C}{4}\right)\sin\left(\frac{C+A}{4}\right) \geq \left(\sin\left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)\right)\left( \sin\left(\frac{B}{2}\right)+\sin\left(\frac{C}{2}\right)\right)\left( \sin\left(\frac{C}{2}\right)+\sin\left(\frac{A}{2}\right)\right)$ Also, using $(a+b)(b+c)(c+a) \geq 8abc$ where $a,b,c \in \mathbb{R^{+}}$ we have
$\left(\sin\left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)\right)\left( \sin\left(\frac{B}{2}\right)+\sin\left(\frac{C}{2}\right)\right)\left( \sin\left(\frac{C}{2}\right)+\sin\left(\frac{A}{2}\right)\right) \geq 8\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)$ Combining the last two inequalities we get the desired claim. Thus, the inradius of triangle $A_{1}B_{1}C_{1}$ is not less than the inradius of triangle $ABC$ . $\blacksquare$ ( $\mathcal{QED}$ )

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