The median $ AM $ and the height $ BH $ of the $ ABC $ triangle ($ H $ on the $ AC $ side) intersect at the $ P $ point. Find $ PH $ if $ AM = BH = 196 $, $ MN = 61 $ where $ N $ is the intersection point of the continuation $AM $ with the circle circumscribed around the triangle $ ABC $. In response, write down the sum of possible values $ PH $.
I can't come out with the solution, I tried. Please anyone help me with the solution. I think I need to form an equation or inequality to find all the possible values.
Since $BH=AM$ and both are opposite $\angle C$, triangles $BHC$ and $AMC$ have same circumradius namely, $R_{\triangle AMC}=MC$.
By sine-rule in $AMC$,
$$\frac{AM}{\sin C}=2R_{\triangle AMC}=\frac{MC}{1/2}=\frac{MC}{\sin 30}$$ $$\therefore \angle MAC=30 \Rightarrow \angle BPM=60$$
If we let $PH=x$, then $AP=2x$, $PM=196-2x$, $BP=196-x$. Also $BM^2=AM\cdot MN=196\cdot 61$
By cosine-rule in $\triangle BPM$,
$$BM^2=BP^2+PM^2-BP\cdot PM$$ $$196\cdot 61=(196-x)^2+(196-2x)^2-(196-2x)(196-x)$$
Sum of values of $PH$ is $(-\,\text{coeff of}\, x/\text{coeff of} \, x^2)$ of this quadratic.