1
$\begingroup$

Let ABC be a triangle with side lengths $AB = 24, CA = 34, BC = 50$. Call the circle with diameter $BC$ ω with O as the center. Let $BA$ intersect $ω$ at $D$ ($D \ne B$) and $CA$ intersect ω at $E$ ($E\ne C$). Let the reflection of $D$ over $O $be defined as $D'$, and have $OA$ and $D'E$ intersect at $F$. Let$ G$ be a point on segment $BC$ such that $CD = 7BG$. Find the length of segment $FG$.

So I have tried to use power of a point with Ptolemy's theorem but doesn't find it works well since there aren't really enough info... Any help on how to do this will be greatly appreciated

EDIT: apparently this problem is really complex (all long geometry problem is complex when there isn't similar triangle! jk) So the simplest way is to use law of sine+cosine, property of cyclic quadrilaterals? Is there another solution to this with less intuition/ calculation? I just think I could never think about adding of point f' which the solution did :(

$\endgroup$
  • $\begingroup$ * Application $\endgroup$ – gen-z ready to perish Sep 21 '17 at 13:10
  • $\begingroup$ Under what circumstances you can apply the power of a point? Is it because FB is tangent to BDCD'? $\endgroup$ – Mick Sep 21 '17 at 14:49
  • $\begingroup$ oh I mean I am applying power of a point at the beginning to get $AE/AD=17/12$ and I tried to use Ptolemy's theorem on both BDCD' and BDCE but they don't work $\endgroup$ – Guy who failed everything Sep 21 '17 at 14:57
  • $\begingroup$ Have you ever thought of $\angle FBO = 90^0$?. If that is true, the whole problem can be solved. $\endgroup$ – Mick Sep 21 '17 at 14:59
  • $\begingroup$ wait how do you get there? is it by angle chasing? $\endgroup$ – Guy who failed everything Sep 21 '17 at 15:09
1
$\begingroup$

Here's a scheme of the solution.

  1. By the cosine rule you can find angles $\angle ABC$ and $\angle ACB$: $$ \cos(\angle ABC)={4\over5},\quad \sin(\angle ABC)={3\over5},\quad \cos(\angle ACB)={154\over170},\quad \sin(\angle ACB)={72\over170}. $$

  2. From that you get $CD=BD'=30$.

  3. Let $F'$ be the point where line $ED'$ meets the tangent at $B$, and apply the sine rule to triangle $BD'F'$. By taking into account that $\angle BD'F'=\angle BCA$ and $\angle F'BD'=\pi/2+\angle DCB=\pi-\angle ABC$, one gets: $\displaystyle BF'={1800\over29}$.

  4. By the cosine rule applied to $BAO$ you can first compute $OA=\sqrt{241}$ and then find $\cos\angle BOA$, from which one obtains $\displaystyle \tan\angle BOA={72\over29}$.

  5. But we also have $\displaystyle {BF'\over OB}={72\over29}$, hence line $OA$ intersects tangent $BF'$ at $F'$, and $F'=F$.

  6. Now you know $BF={1800\over29}$ and $BG={30\over7}$, thus you can find $FG$ by Pythagoras' theorem.

enter image description here

EDIT.

Notice that $OA$ and $D'E$ meet on the line tangent at $B$ not only in this case, but for any position of point $A$ inside the circle of diameter $BC$. I don't know, however, a simple proof for that.

$\endgroup$
  • $\begingroup$ wait i think cos(abc) should be 4/5 though $\endgroup$ – Guy who failed everything Sep 21 '17 at 21:11
  • $\begingroup$ Right, thank you: corrected now. $\endgroup$ – Aretino Sep 21 '17 at 21:12
  • $\begingroup$ and do you mean sinACB= 72/170 instead of sinABC=72/170? $\endgroup$ – Guy who failed everything Sep 21 '17 at 21:15
  • $\begingroup$ Yes, of course: just another typo. $\endgroup$ – Aretino Sep 21 '17 at 21:17
  • $\begingroup$ and can you just show how do you arrive at CD=BD'=30? I tried law of sines but it doesm't work. $\endgroup$ – Guy who failed everything Sep 21 '17 at 21:18
1
$\begingroup$

It looks that Stewart’s Theorem is useful here.

enter image description here

It states that in $\triangle ABC$ \begin{align} a^2\,n+b^2\,m&=c\,(d^2+m\,n) . \end{align}

enter image description here

Fist, some helpful properties precalculated: \begin{align} S_{ABC}&=360 ,\quad \sin\alpha=\tfrac{15}{17} ,\quad \sin\beta=\tfrac35 ,\quad \sin\gamma=\tfrac{36}{85} ,\quad \cos\beta=\tfrac45 ,\quad \cos\gamma=\tfrac{77}{85} ,\\ \end{align}

\begin{align} |CD|&=|BC|\,\sin\beta =30 ,\\ |BG|&=\tfrac17|CD|=\tfrac{30}7 ,\\ |OG|&=\tfrac12|BC|-|BG|=\tfrac{145}{7} ,\\ |BE|&=|BC|\,\sin\gamma =\tfrac{360}{17} ,\\ |BD|&=|BC|\cos\beta=40 ,\\ |CE|&=|BC|\cos\gamma=\tfrac{770}{17} ,\\ |AE|&=|CE|-|AC|=\tfrac{192}{17} ,\\ \angle ED'D &=\tfrac\pi2-\beta-\gamma ,\\ |DE|&=|BC|\,\sin\angle ED'D \\ &=|BC|\,\sin(\tfrac\pi2-\beta-\gamma) \\ &=|BC|\,\cos(\beta+\gamma) \\ &=|BC|\,(\cos\beta\cos\gamma-\sin\beta\sin\gamma) \\ &=|BC|\,\left(\sqrt{(1-\sin^2\beta)(1-\sin^2\gamma)}-\sin\beta\sin\gamma\right) \\ &=|BC|\,\tfrac8{17} =\tfrac{400}{17} ,\\ |D'E|&=\sqrt{|BC|^2-|DE^2|} =\tfrac{750}{17} ,\\ \end{align}

According to Stewart’s Theorem, in $\triangle ABC$

\begin{align} |OA|&= \sqrt{\tfrac12(|AB|^2+|AC|^2)-\tfrac14|BC|^2} =\sqrt{241} ,\\ |AM|&=25-\sqrt{241} . \end{align}

According to Stewart’s Theorem, in $\triangle OME$

\begin{align} |EM|&= \sqrt{\frac{|OM|(|AE|^2+|OA||AM|)-|OE|^2|AM|}{|OA|}} =\tfrac5{4097}\sqrt{839270450-51431810\sqrt{241}} . \end{align}

According to Stewart’s Theorem, in $\triangle ODM$

\begin{align} |MD|&= \sqrt{\frac{|OM|\,(|AD|^2+|OA|\,|AM|)-|OD|^2|AM|}{|OA|}} =\tfrac5{241}\sqrt{2904050-147010\sqrt{241}} . \end{align}

According to Stewart’s Theorem, in $\triangle BOM$

\begin{align} |BM|&= \sqrt{\frac{\tfrac12|BC|(|AB|^2+|OA|\,|AM|)-1/4*BC^2*AM}{|OA|}} =\tfrac5{241}\sqrt{2904050-69890\sqrt{24}} . \end{align}

Also, \begin{align} \cos\angle BOM&= 1-2\sin^2\tfrac12\angle BOM =1-2\left(\frac{|BM|}{|BC|}\right)^2 =\tfrac{29}{1205}\sqrt{241} . \end{align}

According to the power of a point $F$,

\begin{align} \frac{|EF|}{|NF|} &= \frac{|MF|}{|D'F|} =k \tag{1}\label{1} , \end{align}

hence $\triangle MFE \sim \triangle D'FN$ and \begin{align} k& =\frac{|EM|}{|D'N|} =\frac{|EM|}{|MD|} =\tfrac5{102}\sqrt{241}-\tfrac{29}{102} . \end{align}

Now \eqref{1} can be rewritten as a system of two equations in two unknowns, $|EF|$ and $MF$:

\begin{align} \frac{|EF|}{|MF|+|BC|} &=k ,\\ \frac{|MF|}{|EF|+|D'E|}&=k , \end{align}

it follows that \begin{align} |MF|&= \frac{k\,(k\,|BC|+|D'E|)}{1-k^2} =\tfrac{125}{29}\sqrt{241}-25 ,\\ |EF|&= \frac{k\,(k\,|D'E|+|BC|)}{1-k^2} =\tfrac{21600}{493} ,\\ |OF|&=|MF|+\tfrac12|BC|=\tfrac{125}{29}\sqrt{241} . \end{align}

Finally, from $\triangle FGO$, \begin{align} |FG|&= \sqrt{ |OG|^2+|OF|^2-2|OG|\,|OF|\cos\angle BOA } =\frac{12630}{203} . \end{align}

As a bonus, \begin{align} |BF|^2&= \tfrac14|BC|^2+|OF|^2-|BC|\,|OF|\cos\angle BOA =\tfrac{3240000}{841} ,\\ |FG|^2-|BG|^2 &=\tfrac{3240000}{841} , \end{align}

so, indeed, $BF\perp BC$.

$\endgroup$
1
$\begingroup$

As @Mick suggests in a comment, once you know $\overline{FB}\perp\overline{BC}$, the problem is solved, so I'll just prove that. However, instead of constructing $F$ and showing the perpendicularity property, I'll construct the perpendicular at $B$ and show that it concurs with $\overleftrightarrow{OA}$ and $\overrightarrow{D^\prime E}$, via the trigonometric form of Ceva's Theorem.


Consider $\triangle ABE$ in the figure:

enter image description here

In order to name relevant angles, define $A^\prime$ on $\overleftrightarrow{OA}$, $E^\prime$ on $\overleftrightarrow{D^\prime E}$, and $B^\prime$ on the perpendicular to $\overleftrightarrow{BC}$ at $B$, such that $\overrightarrow{AA^\prime}$, $\overrightarrow{BB^\prime}$, $\overrightarrow{EE^\prime}$ are directed toward the ostensible point of concurrency. (Note: Our diagram and argument assume that $\angle A$ is obtuse; equivalently, that $A$ is inside the circle. The reader is invited to make appropriate adjustments for the acute case, where $A$ is outside the circle.)

That concurrency is guaranteed if we can show $$ \frac{\sin\angle A^\prime AE}{\sin\angle A^\prime AB} \; \frac{\sin\angle B^\prime BA}{\sin\angle B^\prime BE} \; \frac{\sin\angle E^\prime EB}{\sin\angle E^\prime EA} = 1 \tag{1}$$

Well, consider the following (where I'll write $\angle B$ and $\angle C$ for the angles at those vertices in $\triangle ABC$) ...

  • $\angle B^\prime BA = 90^\circ - \angle B$, clearly. So, $\sin\angle B^\prime BA = \cos B$.
  • $\angle B^\prime B E = \angle C$, as inscribed angles subtending $\stackrel{\frown}{BE}$. So, $\sin\angle B^\prime BE = \sin C$.
  • $\angle E^\prime E B = 180^\circ - \angle B$. This follows from the fact that $\angle BED^\prime \cong \angle B$ as inscribed angles subtending congruent arcs $\stackrel{\frown}{CD}$ and $\stackrel{\frown}{BD^\prime}$. So $\sin \angle E^\prime E B = \sin B$.

  • $\angle E^\prime EA = 90^\circ +\angle B$. This follows from the additional fact that, by Thales's Theorem, $\angle BEC$ is a right angle. So $\sin \angle E^\prime EA = \cos B$.

Consequently, the left-hand-side of $(1)$ reduces to

$$ \frac{\sin\angle A^\prime AE}{\sin\angle A^\prime AB} \; \frac{\cos B}{\sin C} \; \frac{\sin B}{\cos B} \qquad\to\qquad \frac{\sin\angle A^\prime AE}{\sin\angle A^\prime AB} \; \frac{\sin B}{\sin C} \tag{2}$$

Now, observe that $\angle A^\prime A B$ is the supplement of $\angle OAB$, whereas $\angle A^\prime AE \cong \angle OAC$. Therefore, $(2)$ becomes $$\frac{\sin\angle OAC}{\sin C} \; \frac{\sin B}{\sin\angle OAB} \tag{3}$$

where we've arranged the elements in anticipation of invoking the Law of Sines in $\triangle OAB$ and $\triangle OAC$.

$$\frac{|\overline{OC}|}{|\overline{OA}|} \; \frac{|\overline{OA}|}{|\overline{OB}|} \qquad\to\qquad 1 \tag{4}$$

Thus, the relation is proven, and Ceva guarantees that the perpendicular at $B$ meets the point of intersection ($F$) of $\overleftrightarrow{OA}$ and $\overleftrightarrow{D^\prime E}$, as claimed. $\square$

$\endgroup$
  • $\begingroup$ wait will Pascal's theorem give us the concurrency immediately? $\endgroup$ – Guy who failed everything Sep 23 '17 at 12:05
  • $\begingroup$ @Guywhofailedeverything: I'm not seeing it. $\endgroup$ – Blue Sep 23 '17 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.