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#1 h Apr 18, 2009, 4:51 PM • 1 Y

Y by Adventure10

Show that Euler line of a nonequilateral $\triangle ABC$ is tangent to its incircle $(I),$ if and only if the following relation holds:

$\sqrt {1 - 8\cos A \cos B \cos C} = \frac {| (\sin A - \sin B)(\sin B - \sin C)(\sin C - \sin A) |}{(1 - \cos A)(1 - \cos B )(1 - \cos C)}$

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#2 h May 4, 2009, 12:03 AM • 1 Y

Y by Adventure10

Standard notation: $(O, R)$ is the circumcircle, $(I, r)$ the incircle, $H$ the orthocenter, $a = BC, b = CA, c = AB$ and $p$ is the triangle semiperimeter. In addition, $K, L, M$ are midpoints of $BC, CA, AB;$ $D, E, F$ are incircle tangency points with $BC, CA, AB;$ $U, V, W$ are feet of the altitudes $AH, BH, CH;$ the internal angle bisectors $AI, BI, CI$ cut the circumcircle again at $X, Y, Z.$

Power of $H$ to $(O)$ is $HO^2 - R^2 = 2\ \overline{HU} \cdot \overline{HA} = - 8R^2 \cos A \cos B \cos C$ $\Longrightarrow$ the LHS is equal to $\frac {HO}{R}$ . Using the result of http://www.mathlinks.ro/viewtopic.php?t=256552 (also proved at http://www.mathlinks.ro/viewtopic.php?t=260863), the RHS is equal to

$\frac {|(b - c)(c - a)(a - b)|}{8R^3 \cdot \prod (1 - \cos A)} = \frac {|\overline{KD} \cdot \overline{LE} \cdot \overline{MF}|}{KX \cdot LY \cdot MZ} = \frac {|\overline{DU} \cdot \overline{EV} \cdot \overline{FW}|}{r^3} =$

$= \frac {|(b - c) (p - a) \cdot (c - a)(p - b) \cdot (a - b)(p - c)|}{abc \cdot r^3} =$

$= \frac {|(b - c)(c - a)(a - b)| \cdot |\triangle ABC|^2}{4R \cdot |\triangle ABC| \cdot r^3 \cdot p} = \frac {|(b - c)(c - a)(a - b)|}{4Rr^2}$

$\Longrightarrow$ the given trigonometric condition is equivalent to $HO = \frac {|(b - c)(c - a)(a - b)|}{4r^2}$ . The Euler line $OH$ is tangent to the incircle $\Longleftrightarrow$ $|\triangle IOH| = \frac {_r}{^2}\cdot HO.$ It is therefore sufficient to show that for arbitrary $\triangle ABC,$ $|\triangle IOH| = \frac {|(b - c)(c - a)(a - b)|}{8r}$ . WLOG, assume $a > b > c.$ This area can be calculated from areas of the trapezoids $\square KOHU, \square KOID, \square DIHU.$

$|\triangle IOH| = |\square KOHU| - |\square KOID| - |\square DIHU| =$
$= \frac {_1}{^2}(OK + HU) \cdot KU - \frac {_1}{^2}(OK + ID) \cdot KD - \frac {_1}{^2}(ID + HU) \cdot DU =$
$= \frac {_1}{^2} (OK \cdot DU + HU \cdot KD - ID \cdot KU) =$

$= \frac {_1}{^2} [R \cos A \cdot \frac {(b - c)(p - a)}{a} + 2R \cos B \cos C \cdot \frac {b - c}{2} - R(\cos A + \cos B + \cos C - 1) \cdot \frac {b^2 - c^2}{2a}] =$

$= R\ \frac {b - c}{4a} ((b + c - a) \cos A + 2a \cos B \cos C - (b + c)(\cos A + \cos B + \cos C - 1)) =$

$= R\ \frac {b - c}{4a} (a(2 \cos B \cos C - \cos A) - (b + c)(\cos B + \cos C - 1)).$

Substituting:

(1) $b \cos C + c \cos B = a,$

(2) $R (a \cos A + b \cos B + c \cos C) = 2|\triangle ABC|,$

(3) $2aR (\cos A + \cos B \cos C) = ah_a = 2|\triangle ABC|\ \Longrightarrow$

$|\triangle IOH| = R\ \frac {b - c}{4a} (b + c + a - 2a(1 + \cos A)) =$ $R\ \frac {b - c}{2a} (\frac {|\triangle ABC|}{r} - \frac {a \sin A}{\tan \frac {A}{2}}) =$

${ = \frac {b - c}{8r} (bc - 2a(p - a)}) = \frac {|(b - c)(c - a)(a - b)|}{8r}$

EDIT: I should have searched earlier. See http://www.mathlinks.ro/viewtopic.php?t=2878.

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#3 h May 4, 2009, 2:22 AM • 1 Y

Y by Adventure10

Thanks for your very nice solution yetti!. I also used the result of the topic An IOH smart and interesting problem. Incircle $(I,r)$ is tangent to the Euler line $\Longleftrightarrow$ $| \triangle IOH| = \frac {_1}{^2}r \cdot OH.$ Hence

$\frac {r^2 \cdot OH^2}{4} = | \triangle IOH|^2 = \frac {(a - b)^2(b - c)^2(c - a)^2}{64r^2}$

Now, we use $OH^2 = 9R^2 - (a^2 + b^2 + c^2),$ together with the identities

$a = 2R \cdot \sin A \ , \ b = 2R \cdot \sin B \ , \ c = 2R \cdot \sin C$

$\frac {9}{4} - \sum \sin^2 A = \frac {R^4}{r^4}(\sin A - \sin B)^2 \cdot (\sin B - \sin C)^2 \cdot (\sin C - \sin A)^2 =$

$\frac {1}{4} - 2\cos A \cdot \cos B \cdot \cos C = \frac {R^4}{r^4}(\sin A - \sin B)^2 (\sin B - \sin C)^2 (\sin C - \sin A)^2$

In the $\text{RHS}$ we use the identity $\frac {r}{R} = 4\sqrt {\frac {_1}{^8}(1 - \cos A)(1- \cos B)(1-\cos C)}$

After extracting the square root in both sides of the equation, we obtain

$\sqrt {1 - 8\cos A \cos B \cos C} = \frac {| (\sin A - \sin B)(\sin B - \sin C)(\sin C - \sin A)|}{(1-\cos A)(1-\cos B) (1 - \cos C)}$

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