Sum of product of distances is constant

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Sum of product of distances is constant

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Source: Bundeswettbewerb Mathematik 2020, Round 1 - Problem 3

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#1 h Nov 17, 2020, 3:31 PM • 2 Y

Y by Mango247, Mango247

Let $AB$ be the diameter of a circle $k$ and let $E$ be a point in the interior of $k$ . The line $AE$ intersects $k$ a second time in $C \ne A$ and the line $BE$ intersects $k$ a second time in $D \ne B$ .

Show that the value of $AC \cdot AE+BD\cdot BE$ is independent of the choice of $E$ .

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#2 h Nov 17, 2020, 6:24 PM

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Let $EG \perp AB$ then $AC*AE= AB \cos \angle CAB * \frac{AG}{\cos \angle CAB}= AB*AG$
$BD*DE=AB*BG$
$AC \cdot AE+BD\cdot BE=AB^2$

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#4 h Mar 23, 2021, 12:57 PM

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Quick trig solution for storage

Obviously $\angle ACB=\angle ADB=90^\circ$ , so from triangle $ABC$ we have: $AC=AB\cdot\cos\angle BAE$ and from triangle $ABD$ we have: $BD=AB\cdot\cos\angle ABE$ . $\angle AEB=180^\circ-\angle ABE-\angle BAE$ , so $\sin\angle AEB=\sin(\angle BAE+\angle ABE)$ . Now from sine law in triangle $ABE:$
$BE=\frac{AB\cdot\sin\angle BAE}{\sin(\angle BAE+\angle ABE)}$ and $AE=\frac{AB\cdot\sin\angle ABE}{\sin(\angle BAE+\angle ABE)}$ , so finally:
$AC \cdot AE+BD\cdot BE=AB\cdot\cos\angle BAE\cdot\frac{AB\cdot\sin\angle ABE}{\sin(\angle BAE+\angle ABE)}+AB\cdot\cos\angle ABE\cdot\frac{AB\cdot\sin\angle BAE}{\sin(\angle BAE+\angle ABE)}=$
$AB^2(\frac{\cos\angle BAE\cdot\sin\angle ABE+\cos\angle ABE\cdot\sin\angle BAE}{\sin(\angle BAE+\angle ABE)})=AB^2\frac{\sin(\angle BAE+\angle ABE)}{\sin(\angle BAE+\angle ABE)}=AB^2$
So it does not depend on choosing point $E$ .

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