Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$|AM|=|CM|$

$\angle BCA = 15^{\circ}$

$\angle CBM = \angle ABH$

$\angle BHC = 90^\circ$

Find $|AC|$

enter image description here

The solution states that $\overline{BM}$ is the isogonal conjugate of $\overline{BH}$ but I did not understand this at all. Could you explain this to me or maybe find another way to solve it?

share|improve this question
    
Isogonal conjugate just means that $\angle CBM = \angle ABH$. There are some properties which may come into play later, but you should be able to read on. –  Calvin Lin Feb 4 at 22:14
    
To check, are you also given that $\angle CBA = 90^\circ$? Otherwise it seems like there could be multiple angles, just by scaling to fit the length of $MH$. –  Calvin Lin Feb 4 at 22:17
    
The solution says that because the median $\overline{BM}$ is the isogonal conjugate of the altitude $\overline{BH}$, $\angle CBA = 90^{\circ}$ but I don't understand that. –  Zafer Cesur Feb 4 at 22:26
    
@CalvinLin It seems he's given $\angle BHC = 90^\circ$, from the drawing. Using that, experimentation in GeoGebra seems to suggest that there is only one case (where $\triangle BMH$ was non-degenerate). –  Arthur Feb 4 at 22:26
    
Oh, yes. $\angle BHC = 90^\circ$ is given. –  Zafer Cesur Feb 4 at 22:27

2 Answers 2

up vote 1 down vote accepted

Call the angle at $A$ $\alpha$ and let $O$ be the circumcenter.

Clearly, $\measuredangle HBA = 90-\alpha$.

Now, by the inscribed angle theorem: $\measuredangle CBO = \frac{180-\measuredangle BOC}2 =90-\measuredangle BAC = 90- \alpha$

Therefore, both $M$ and $O$ lie on the line $BM$ as well as on the perpendicular bisector of $CA$. This means that either the triangle isosceles or $M=O$. In the first case, $M=H$ which is clearly impossible from the given distance.

However, if $M=O$, the triangle has be orthogonal at $B$. The similarity of $ABC$ and $AHB$ now permits to calculate all angles.

You can finally use the sine law and the given distance to calculate $BH$, $AH$ and $CH$ which gives you $AC$.

share|improve this answer

$\overline{BM}$ is the Isogonal conjugate of $\overline{BH}$ means BM is pass through the circumcentre as $\overline{BH}$ is pass the orthocentre. so $M$ is circumcentre $\implies \angle CBA= 90^\circ $. now you can find every thing.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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