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I recently came across a geometry problem , published in an local magazine(publishing at high school and under graduate level) and was under Difficulty : Hard sub heading.

Consider a $\triangle ABC$ , then $D$ is a point on $BC$ such that , if the circumcircle of $\triangle ABD$ is drawn and the tangent drawn from point $C$ to the Circumcircle touches the circle at $E$ then : $\angle DCE= \angle DAC.$ It is given that $\frac{5}{3}BC=AB+AC$ and $AC=33 \text{ units}$. Then find the value of $CD$ . A not to scale and totally rough figure of it has been provided here by me.

I tried to work out and found mutiple angles to be congruent hinting similarity also the mention of $\frac{5}{3}$ in the question releases a strong smell of Similarity of triangles!

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  • $\begingroup$ @TedShifrin sorry ...I will correct it in the figure. $\endgroup$ – Shivam Patel Jul 2 '16 at 17:30
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    $\begingroup$ A good starting point is to notice that $CE$ is tangent to the circumcircle of $CAD$, too. $\endgroup$ – Jack D'Aurizio Jul 2 '16 at 17:51
  • $\begingroup$ I have a quite involved solution through complex numbers, but I hope to find something better (more elementary) soon. $\endgroup$ – Jack D'Aurizio Jul 2 '16 at 18:46
  • $\begingroup$ @Jack D'Aurizio You can post the complex number solution and then add the elementary one... Because it would be really great for me to see the problem tackled in a different light $\endgroup$ – Shivam Patel Jul 3 '16 at 14:30
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enter image description here

The angle constraint gives that the circumcircle of $CDA$ is also tangent to the $CE$-line.
So the three quantities $d=CE$, $R=OE$, $\theta=\widehat{BCE}$ fix the above configuration: the circumcentre of $CDA$ is given by the intersection between the perpendicular bisector of $CD$ and the perpendicular to $CE$ through $C$. Now it is just a terribly painful trigonometry exercise to describe the lengths of $CD,DB,BA,AC$ in terms of $R,d,\theta$. Do that, then impose the constraints $AC=33$ and $BC=\frac{3}{5}AB+\frac{3}{5}AC$ to find the length of $CD$: the wanted configuration is reached at $\widehat{CBA}=90^\circ$.

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  • $\begingroup$ Just curious, how did you draw the image? Using Paint or something? $\endgroup$ – Babai Jul 11 '16 at 4:58

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