10th ibmo - chile 1995/q3.

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10th ibmo - chile 1995/q3.

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similar triangles

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Source: September 23rd - 30th

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#1 h May 7, 2006, 1:48 PM • 2 Y

Y by Adventure10, Mango247

Let $r$ and $s$ two orthogonal lines that does not lay on the same plane. Let $AB$ be their common perpendicular, where $A\in{}r$ and $B\in{}s$ (*).Consider the sphere of diameter $AB$ . The points $M\in{r}$ and $N\in{s}$ varies with the condition that $MN$ is tangent to the sphere on the point $T$ . Find the locus of $T$ .

Note: The plane that contains $B$ and $r$ is perpendicular to $s$ .

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#2 h Sep 8, 2009, 6:07 PM • 2 Y

Y by Adventure10, Mango247

Choose an orthogonal system of coordinates such that the center of the sphere with radius $R$ is $O = (0,0,0)$ , $A = (0,0,R)$ , $B = (0,0, - R)$ . Clearly nonzero reals $m,n$ exists such that $M = (0,m,R)$ and $N = (n,0,R)$ , and line $MN$ is given by $2Rx + nz - nR = 0$ , $2Ry - mz - mR = 0$ . Replace the values of $x,y$ as a function of $z$ in the equation of the sphere $x^2 + y^2 + z^2 = R^2$ , obtain a 2nd degree equation in $z$ . Since $MN$ must be tangent to the sphere, the discriminant must be zero, obtain from here condition $|mn| = 2R^2$ and solution $z = - R\frac {m^2 - n^2}{m^2 + n^2 + 4R^2}$ for the point of tangency. Substitute in the equations for $MN$ , use $|mn| = 2R^2$ , and find that the point of tangency satisfies $x = \frac {n|m|}{|m| + |n|}$ , $y = \frac {m|n|}{|m| + |n|}$ , or $|x| = |y|$ . Hence $T$ is necessarily a point in the meridians that result from the intersection of the surface of the sphere with the planes that bisect the dihedron formed by the planes that contain $AB$ and $r$ , and $AB$ and $s$ , except for the poles $A,B$ themselves. Take any point in these meridians, do the construction backward so all these points are in fact the locus of $T$ .

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#3 h Jan 14, 2010, 4:37 PM • 2 Y

Y by Adventure10, Mango247

Let $\alpha , \beta$ denote the planes determined by $AB , s$ and $AB , r$ and let $X,Y$ be the orthogonal projections of $T$ on $\alpha$ and $\beta.$ Since $X, Y$ lie on the orthogonal projection of $MN$ on $\alpha, \beta,$ we get the similar triangles $\triangle TXN \sim \triangle MAN$ and $\triangle TYM \sim \triangle NBM$ $\Longrightarrow$

$\frac {\overline{TX}}{\overline{MA}} = \frac {\overline{NT}}{\overline{MN}} \ , \ \frac {\overline{TY}}{\overline{NB}} = \frac {\overline{MT}}{\overline{MN}}$

Since obviously $\overline{NT} = \overline{NB}$ and $\overline{MA} = \overline{MT},$ we deduce that $\overline{TX} = \overline{TY}$

This means that $T$ lies on some of the two angle bisector (planes) $\gamma$ and $\gamma'$ determined by the dihedral angle $\alpha\beta.$ Therefore, loci of $T$ are two circumferences given from the intersection of the sphere with both planes $\gamma , \gamma'.$

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