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Let $AB,CD$ be two lines inscribed in a circle that intersect each other at $E$($E$ is in the circle).Let $M$ be a point on $EB$ between $E$ and $B$.Let $C(O)$ be a circle that passes from three points $M,D,E$ we draw a tangent to this circle at point $E$ that intersects $BC,AC$ at $F,G$.If $\frac{AM}{AB}=t$ then find $\frac{EG}{EF}$ in terms of $t$.

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Because our lesson was about similarities I tried to use the tangetnt with angles to find a smilarity.Instead of finding some smilarities they didn't help me to find $\frac{EG}{EF}$ in terms of $t$.

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    $\begingroup$ If $M$ is taken at random on $EB$, then we should disregard the apparent tangency of the circles at $D$? $\endgroup$ – Edward Porcella Nov 16 '17 at 19:55
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AM/AB in terms of tStarting with the posted figure, through $A$ draw $AHK$ parallel to $GEF$, and join $HM$. Then $$\frac{GE}{EF}=\frac{AH}{HK}$$ Further, $\angle ADE=\angle HME$, owing to the similarity of triangles $ADE$ and $HME$. This similarity follows from the fact, not here proven, that $H$ is concyclic with $A$, $D$, $M$. (For an easy way to show it, see @Aretino comment.)

But $\angle ADE = \angle CBA$, since they stand on arc $AC$.

Therefore, $\angle HME=\angle CBM$, making $HM$ parallel to $CB$.

Hence $$\frac{AH}{HK}=\frac{AM}{MB}$$

Therefore, $$\frac{GE}{EF}=\frac{AM}{MB}$$

Since $\frac{AM}{AB}=t$, then we have$$\frac{GE}{GE+EF}=\frac{AM}{AM+MB}=\frac{AM}{AB}=t$$

From this it follows that$$\frac{EG}{EF}=\frac{t}{1-t}$$

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    $\begingroup$ To show that $HADM$ are concyclic, just notice that $\angle EDM=\angle FEM=\angle HAM$. $\endgroup$ – Aretino Nov 17 '17 at 10:55
  • $\begingroup$ @Aretino--Thank you. Euclid III, 32: I tried earlier to get some use of it but without success. Here it's just the thing. $\endgroup$ – Edward Porcella Nov 17 '17 at 16:40
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$$\frac{EG}{EF}=\frac{t}{1-t}$$

I did a massive coordinate-based brute-force computation, based on some concepts from projective geometry. Without loss of generality $A,B,C,D$ are points on the unit circle, none of them $(-1,0)$. Start with four parameters $a,b,c,d$ and use them to describe four points on the unit circle using the tangent half-angle formula. Use homogeneous coordinates to describe these.

$$A=\begin{pmatrix}1-a^2\\2a\\1+a^2\end{pmatrix}\qquad B=\begin{pmatrix}1-b^2\\2b\\1+b^2\end{pmatrix}\qquad C=\begin{pmatrix}1-c^2\\2c\\1+c^2\end{pmatrix}\qquad D=\begin{pmatrix}1-d^2\\2d\\1+d^2\end{pmatrix}$$

Compute $E$ as

$$E=(A\times B)\times (C\times D)$$

Compute $M$ as

$$M=(1-t)B_3A + tA_3B$$

scaling by the third coordinate to obtain compatible homogeneous representatives. Find the matrix of the circle passing through $D,E,M$ to be

$$Q=\scriptscriptstyle\begin{pmatrix} - a b c - a b d + a c d + b c d - a - b + c + d & 0 & - a c d t + b c d t - a b d + a c d + a t - b t + b - c \\ 0 & - a b c - a b d + a c d + b c d - a - b + c + d & a c t - b c t + a d t - b d t + a b - a c + b d - c d \\ - a c d t + b c d t - a b d + a c d + a t - b t + b - c & a c t - b c t + a d t - b d t + a b - a c + b d - c d & -2 a c d t + 2 b c d t + a b c - a b d + a c d - b c d - 2 a t + 2 b t + a - b + c - d \end{pmatrix}$$

One way to do this is to find the conic through five points, the three named above and the ideal circle points $(1,\pm i,0)$. Details might be best served in a separate question. Suffice it to show that

$$0=D^T\cdot Q\cdot D=E^T\cdot Q\cdot E=M^T\cdot Q\cdot M$$

so all three points are incident with the circle. Now

$$g=Q\cdot E$$

is the tangent to this circle in point $E$. This is a special case of computing the polar line to a given point using this conic times point form. Then

$$F=g\times(B\times C)\qquad G=g\times(A\times C)$$

Dehomogenize the points (i.e. divide by third coordinate) and compute distance vectors. As I tried to do all the computations using polynomials and avoid square roots, I computed the squares of the norms of the corresponding distances.

$$\frac{\lVert E/E_3-G/G_3\rVert^2}{\lVert E/E_3-F/F_3\rVert^2}= \frac{t^2}{(1-t)^2}$$

Quick sanity check to make sure that the sign makes sense after dropping the squares, and there is the answer. Not an elegant approach, but one that works. And perhaps knowing the desired result will help someone else to find a nicer way to prove it.

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