contest-matheuclidean-geometrygeometrytrianglestrigonometry
We have triangle $\triangle ABC$, with $AD\perp BC$, $\angle ABH=20^o$, $\angle HBC=40^o$ and $\angle HCB=30^o$. Find the value of angle $\angle HAC$
We have that $\angle HAB=30^o$, $\angle BAD=30^o$, $\angle BHD=50^o$, $\angle DHC=60^o$, $\angle AHC=120^o$. After I drew it out accurately, I worked out that the answer is $20^o$. But I can't work out how to show it. Could you please explain to me how to solve this question?
Before solving the problem, let's put ourselves in this situation (see attached image)
If $HD=DJ, HG = 2(GO)$ and $JO=OI$ then you have to necessarily $J, O$ and $I$ are collinear.
A simple test is by contradiction. Suppose they are not collinear, so let's place $L$ in $HO$ such that $DL // JO$ then we would have $HL = 3a$, $LG = a$ and $GO = 2a$ ($a$ is a constant), in addition $DL = 2 (JO) = 2 (OI)$. With all this, we would have that $DL // OI$ (the triangles $DLG$ and $GOI$ are similar). Let's call $\theta = \angle JOL$, then $\angle DLH = \theta$ and $ \angle DLO = \angle LOI = 180 - \theta$, therefore $J,O,I$ are collinear.
In the problem, let's call $H$ the orthocenter of the triangle $ABC$ and $O$ to its circumcenter. It is known that $H,G,O$ are collinear and $HG = 2GO$. Now let $J$ be the intersection of the extension of $AH$ with the circumscribed circumference, so it is easy to see that $HD = DJ$. Finally, we would have by the initial observation that necessarily J, O, I are collinear. Then $G$ is the centroid of the triangle $HIJ$ and $GI=2DG$.
Sorry for having changed the letter $E$ for the $I$, I hope it is understood in the same way
Well, afterall this is an IMO 2001 geometry problem and you can find two solutions, along with a solutions of the other problems from that year, here:
Best Answer
As we discussed in the comments, $H$ is the orthocenter and then simple angle chasing gives $\angle HAC = 40^0$.
Alternatively, by Trigonometric form of Ceva's theorem, we have
$ \sin (\angle HBC) \ \sin (\angle HCA) \ \sin (\angle HAB) = \sin (\angle ABH) \ \sin (\angle HCB) \ \sin (\angle HAC)$
$\sin 40^0 \sin (60^0-x) \sin 30^0 = \sin 20^0 \sin 30^0 \sin x \ $ ($\angle HAC = x)$
$\sin 40^0 \sin (60^0-x) = \sin 20^0 \sin x$
$\cos (20^0 - x) - \cos(100^0 - x) = \cos (20^0-x) - \cos (20^0+x)$
$\implies x = 40^0$