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Consider the following problem.

The Problem

Point $B$ lies on the segment $AC$. A tangent line is constructed from point $A$ to the circle of diameter $BC$, intersecting it at point $M$. $K$ is the second point of intersection of line $AM$ with the circle of diameter $AB$. The extension of segment $MB$ intersects the circle of diameter $AB$ at point $D$.

a) Prove $AD\parallel MC$.

b) Find the area of $DBC$ if $AK=5$ and $MK=25$.

My Drawing

The drawing to the problem

My Solution

a) $\angle ADB = \angle BMC = 90^\circ$ as inscribed angles subtended by the diameter are always right. As $\angle ADB$ and $\angle BMC$ are alternate interior angles of a transversal intersecting two lines $AD$ and $MC$, and as they are equal, then, by Proposition 27 of Euclid (one of criterion for parallel lines) these lines are parallel, i.e. $AD\parallel MC$, QED.

b) Because $AD\parallel MC$, $DAMC$ is a trapezoid with bases $AD$ and $MC$.

As per properties of a trapezoid, its diagonals divide its interior region into four triangles: two triangles that include one of this trapezoid's bases are similar, and two other triangles have the same area. Thus, to find the area of triangle $_\Delta DBC$, it's sufficient to find the area of a triangle $_\Delta ABM$, as their areas are equal.

In $_\Delta ABM$, we can immediately find side $AM=5+25=30$. We now find the altitude to that side in order to find out the area of this triangle. This altitude is $BK$, as $\angle AKB = 90^\circ$ as an inscribed angle subtended by the diameter.

Let $\angle AMB = \alpha$. This is the angle between tangent $AM$ and chord $MB$, thus, by the alternate segment theorem, it is equal to the angle in the alternate segment $BM$. Thus, $\angle BCM = \angle AMB = \alpha$. Then $\angle CBM=90^\circ - \alpha$ as triangle $CMB$ is a right triangle (established in a) earlier), and sum of acute angles in a right triangle is always $90^\circ$.

Now, $\angle KMB=\angle ABM$ because point $K$ lies on a line $AM$ between points $A$ and $M$. Therefore, $\angle KBM=\angle CBM=\alpha$. Because triangle $MKB$ is also a right triangle, then $\angle MBK=90^\circ-\alpha=\angle CBM$. Thus, by axiom of angle measurement, $\angle CBK=\angle MBK+\angle CBM=90^\circ-\alpha+90^\circ-\alpha=180^\circ-2\alpha$. Now, as $\angle ABK$ is adjacent to $\angle CBK$, it's true that $\angle ABK+\angle CBK=180^\circ$. Therefore, $\angle ABK=2\alpha$.

As triangle $AKB$ is a right triangle as $BK\perp AK$, it's true that $\angle BAK=90^\circ-2\alpha$.

Let's now express $\tan\alpha$ in terms of some sides. By definition of a tangent of an acute angle of a right triangle, it's the ratio of the opposite leg to the adjacent leg. Therefore, from triangle $MKB$, $\tan\alpha=\dfrac{BK}{MK}=\dfrac{BK}{25}$.

By one of the phase shift identities, $\tan(90^\circ - 2\alpha)=\cot 2\alpha$. By the double-angle identity for cotangent, $\cot 2\alpha=\dfrac{1-\tan^2\alpha}{2\tan\alpha}$. From the right triangle $ABK$ and the definition of tangent, we also have $\tan(90^\circ - 2\alpha)=\tan\angle BAK=\dfrac{BK}{AK}=\dfrac{BK}{5}$. Therefore, it's true that $\cot 2\alpha=\dfrac{BK}{5}$.

Let $\tan\alpha = x$, and $BK = y$. We now solve the system of equations from what we have got earlier:

$$ \begin{cases} x=\dfrac{y}{25}, \\ \dfrac{1-x^2}{2x} = \dfrac{y}{5}; \end{cases} \Rightarrow \begin{cases} y=25x, \\ y=5\cdot\dfrac{1-x^2}{2x}; \end{cases} \Rightarrow 25x = 5\cdot\dfrac{1-x^2}{2x}. $$

We now solve the equation for $x$:

\begin{align*} 25x &= 5\cdot\dfrac{1-x^2}{2x} \\ 5x &= \dfrac{1-x^2}{2x} \\ 10x^2 &= 1-x^2 \\ 11x^2 &= 1 \\ x^2 &= \dfrac{1}{11} \end{align*}

That means $\tan^2\alpha=\dfrac{1}{11}$. We know $\alpha$ is an acute angle, so $\tan\alpha > 0$. Then, we will take the positive root: $\tan\alpha=\dfrac{1}{\sqrt{11}}$.

Now we can find $BK$.

\begin{align*} \tan\alpha &= \dfrac{BK}{25} \\ \dfrac{BK}{25} &= \dfrac{1}{\sqrt{11}} \\ BK &= \dfrac{25}{\sqrt{11}} \end{align*}

We can now find the area of a triangle $_\Delta ABM$:

$$ S_{_\Delta ABM} = \dfrac{1}{2}\cdot AM\cdot BK = \dfrac{1}{2}\cdot 30\cdot \dfrac{25}{\sqrt{11}} = \dfrac{375}{\sqrt{11}} $$

Therefore, as $S_{_\Delta ABM}=S_{_\Delta DBC}$, we conclude that $S_{_\Delta DBC}=\dfrac{375}{\sqrt{11}}$.

My Question

Is my solution correct, or does it have a flaw? The other person got an answer $\dfrac{75\sqrt{11}}{2}$. Which answer is correct, and why?

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  • $\begingroup$ Your solution is correct. $\endgroup$ Commented Jun 5, 2023 at 15:17
  • $\begingroup$ @Rusuano, you figure does not match the statement, please change it. $\endgroup$
    – sirous
    Commented Jun 6, 2023 at 5:00
  • $\begingroup$ @sirous I'll redo the drawing in GeoGebra in order to meet the quality guidelines. $\endgroup$
    – Rusurano
    Commented Jun 6, 2023 at 7:12

1 Answer 1

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I believe your answer is correct.

I figured this approach is a bit faster, and confirms your result.

Let $O$ be the midpoint of $BC$. Angle chasing yields $MO\parallel KB$ (I leave you this part as an exercise). So, if $\overline{BC} = 2x$, $\overline{AB} = \frac15 x$, and $\overline{KB} = \frac16 x$.

Pythagorean Theorem on $\triangle AKB$ gives $$\frac1{25}x^2 = 25+\frac1{36}x^2,$$ that is $$x=\frac{150}{\sqrt{11}}.$$

Again Pythagoras on $\triangle KBM$ and $\triangle BCM$ yields $$\overline{MB} =\frac{50\sqrt{33}}{11},$$ and $$\overline{MC} = 50\sqrt 3 $$ Similarity $\triangle ADM \sim \triangle MBC$ gives $$\overline{DB} = \frac{5\sqrt{33}}{11}$$ and $$\overline{AD} = 5\sqrt 3.$$ If $DH$ is the line segment perpendicular to $AB$ we get $$\overline{DH} = \frac{\overline{AD}\cdot\overline{DB}}{\overline{AB}}.$$ Thus $$\mathcal A_{DBC} = \frac{\overline{BC}\cdot\overline{AD}\cdot\overline{DB}}{2\overline{AB}}=\frac{375}{\sqrt{11}}.$$

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    $\begingroup$ The 'angle chasing' you've been mentioning is likely this: let $\angle KMB=\alpha$, then by the alternate segment theorem, $\angle KMB=\angle MCB=\alpha$. $MO$ will be the median of hypotenuse $BC$, thus $OM=OB=OC$, which yields $\angle OCM = \angle OMC = \alpha$. Therefore, $\angle OMB=90^\circ - \alpha$, and finally, $\angle OMK=90^\circ-\alpha+\alpha=90^\circ$, which leads to $OM\perp AM$, and since $BK\perp AM$ too, then $OM\parallel BK$, QED. $\endgroup$
    – Rusurano
    Commented Jun 6, 2023 at 7:24

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