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Two circles ...

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Source: INMO 1996 Problem 2

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1592 posts

Rushil

#1 h Oct 6, 2005, 12:33 PM • 3 Y

Y by Adventure10, Mango247, Mango247

Let $C_1$ and $C_2$ be two concentric circles in the plane with radii $R$ and $3R$ respectively. Show that the orthocenter of any triangle inscribed in circle $C_1$ lies in the interior of circle $C_2$ . Conversely, show that every point in the interior of $C_2$ is the orthocenter of some triangle inscribed in $C_1$ .

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3660 posts

Arne

#2 h Oct 7, 2005, 1:16 PM • 2 Y

Y by Adventure10, Mango247

Let $O$ , the center of $C_1$ and $C_2$ , be the origin - I am going to use vectors. Let $ABC$ be any triangle inscribed in $C_1$ and write $\mathbf{a} = \overline{OA}$ , $\mathbf{b} = \overline{OB}$ and $\mathbf{c} = \overline{OC}$ . Then $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = R$ . Also, the orthocentre $H$ of $\Delta ABC$ is given by $\mathbf{h} = \overline{OH} = \mathbf{a} + \mathbf{b} + \mathbf{c}$ . By the triangle inequality, we have $|\mathbf{h}| = |\mathbf{a} + \mathbf{b} + \mathbf{c}| \leq |\mathbf{a}| + |\mathbf{b}| + |\mathbf{c}|= 3R,$ and it follows that $H$ lies inside the circle with centre $O$ and radius $3R$ .

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572 posts

jensen

#3 h Oct 7, 2005, 8:19 PM • 2 Y

Y by Adventure10, Mango247

nice solution Arne,you are cool.

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904 posts

#4 h Oct 3, 2009, 6:39 PM • 2 Y

Y by Adventure10, Mango247

Anyone for pure geometry or coordinate geometry?
Seems quite tough .....
The problem with of is that i havent yet learnt vector use in geometry.........

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904 posts

#5 h Oct 11, 2009, 8:39 AM • 2 Y

Y by Adventure10, Mango247

After the last post , I realised that a very short and trivial solution exists using trigonometry.....

It is a well known relation that

$OH = R \sqrt{1-8cosAcosBcosC}$

where O is the circumcentre of a triangle, H is the orthocentre, R is the circumradius, and A,B,C the angles of the triangle.

Since $cosA \cdot cosB \cdot cosC= cos(180-(B+C)) \cdot cos B \cdot =-cos(B+C) \cdot cosB \cdot cosC < -1$ since the maximumum value of the cos function is $1$ which exists at angle $0$ not possible....

Therefore
$OH = R \sqrt{1-8cosAcosBcosC} < R \sqrt{1+8} = 3R$

This proves the first part.....

For the second part....
Since $cosAcosBcosC$ assumes every value between $-1$ and $\frac{1}{8}$ with equality existing only in the latter.....
Therefore every point in the interior of $C_2$ is the orthocenter of some triangle inscribed in $C_1$ .

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904 posts

#6 h Oct 20, 2009, 5:16 PM • 1 Y

Y by Adventure10

It is a well known relation that

$OH = R \sqrt {1 - 8cosAcosBcosC}$

where O is the circumcentre of a triangle, H is the orthocentre, R is the circumradius, and A,B,C the angles of the triangle.

$cosA \cdot cosB \cdot cosC = cos(180 - (B + C)) \cdot cos B \cdot cos C \\ = - cos(B + C) \cdot cosB \cdot cosC > - 1$
since the maximum value of the cos function is $1$ which exists at angle $0$ not possible....

Therefore
$OH = R \sqrt {1 - 8cosAcosBcosC} < R \sqrt {1 + 8} = 3R$

This proves the first part.....

For the second part....
Since $cosAcosBcosC$ assumes every value between $- 1$ and $\frac {1}{8}$ with equality existing only in the latter.....
Therefore every point on a radius of the circle is the orthocentre of some triangle in $C_1$ ...So rotating those triangles we get the required statement of the proof...
Therefore every point in the interior of $C_2$ is the orthocenter of some triangle inscribed in $C_1$ .

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124 posts

#7 h Oct 30, 2011, 10:10 AM • 3 Y

Y by SahilAgrawal753, Adventure10, Mango247

Here's a solution using pure geometry for the first part:
Use the relation OH = 3OG, and the fact that G (the centroid) must lie inside the triangle, and hence inside the circle. It follows that H must lie inside the circle of radius 3R

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1459 posts

#8 h Nov 4, 2011, 1:42 AM • 1 Y

Y by Adventure10

My first part is same as Pascal96's solution.

For the second part (without trigo), let $O$ be the center of both circles, and $HO$ intersects $C_1$ at $A$ and $X$ , where $AH\leq AX$ . Note that the midpoint of $HX$ lies in $C_1$ , so the perpendicular bisector of $HX$ intersects $C_1$ at 2 points $B,C$ . Then $H$ is the orthocenter of $\triangle ABC$ , not hard to show.

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147 posts

#9 h Dec 2, 2011, 6:41 AM • 1 Y

Y by Adventure10

another proof for the second part:
Let $C_1$ be the unit circle in Argand Plane and the let the positive x-axis be along $OH$ where $O$ is the origin and $H$ is an arbitrary point inside $C_2$ .We have $h=|OH|=k$ (say).Using the fact $h=a+b+c$ it suffices to prove that k can be written as $e^{ix}+e^{iy}+e^{iz}$ .This is equivalent to $\cos x+\cos y+\cos z=k$ and $\sin x+\sin y+\sin z=0$ Choosing $\cos x=\cos y=\frac{k-1}{2}$ (this is possible because $k<3$ ), $\cos z=1$ and $x=-y$ we get a triangle.
$QED$

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452 posts

aops29

#10 h Jan 5, 2020, 9:23 AM • 2 Y

Y by AlastorMoody, Adventure10

Solution

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