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Source: Estonia IMO TST 1996 Day2 P2

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IstekOlympiadTeam

#1 h Nov 27, 2015, 8:00 PM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, Mango247

Let $H$ be the orthocenter of an obtuse triangle $ABC$ and $A_1B_1C_1$ arbitrary points on the sides $BC,AC,AB$ respectively.Prove that the tangents drawn from $H$ to the circles with diametrs $AA_1,BB_1,CC_1$ are equal.

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#2 h Nov 27, 2015, 8:14 PM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, Mango247

Let $B_2, C_2 , A_2$ the intersection of the altitudes with the respective side.
Looking at the triangle $BB_2B_1$ we will notice that $\angle BB_2B_1=90 \implies B_2$ is on the circle with diameter $BB_1$ .
If $HP$ is the tangent from $H$ to this circle we need to prove that $HP^2=HB \cdot HB_2$ is equal to the others.
$\leftrightarrow HB \cdot HB_2=HA \cdot HA_2=HC \cdot HC_2$ which is obvious using the fact that $HA, HB, HC$ are altitudes and they form inscribed quadrilaterals ( $BB_2C_2C, BB_2A_2A , AA_2C_2C$ )

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#3 h Nov 27, 2015, 8:42 PM • 1 Y

Y by Adventure10

Hi,
Proving that the tangents to the circles are equal is equivalent to proving that the power of $H$ wrt to the circles are all equal.
Since $A_1,B_1,C_1$ are arbitrary points on the sides, by fixing one point say $C_1$ we can prove that for all points $A_{1i}$ the power of $H$ wrt to the circle with diameter $AA_{1i}$ is a constant and then for a specific point $A_1$ we prove that $p(H,k_{AA_1})=p(H,k_{CC_1})$ . It goes something like this.
$p(H,k_{AA_1})=HS_a^2- \frac{AA1^2}{4}$ , where $S_a$ is the midpoint of $AA_1$ . By law of cosines $HS_a^2- \frac{AA1^2}{4}=HA^2+\frac{AA1^2}{4}-2HA\frac{AA1}{2} cos( \angle HAA_1)- \frac{AA1^2}{4}=HA(HA-AA_1cos(\angle HAA_1))=HA(HA+AA_1cos(\angle A_1AA_0))$ where $A_0$ is the foot of the altitude from $A$ to $BC$ . Then we have $p(H,k_{AA_1})=HA \times HA_0=const$ . Because $CC_0AA_0$ is cyclic we get that $HA \times HA_0=CH \times CH_0$ so we have $p(H,k_{AA_{1}})=p(H,k_{AA_0})=p(H,k_{CC_0})=p(H,k_{CC_1})$ .
P.S. during my writing of the solutions I got the notification that someone else already posted but nevertheless I finished what I started.

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#4 h Dec 25, 2018, 11:31 AM • 3 Y

Y by wu2481632, Adventure10, Mango247

A much better to prove that $H$ is the radical center of the circles (with the cevians as diameters) is found in Coxeter's Geometry Revisited, page 37:

Solution

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