contest-mathgeometryplane-geometrytriangles
SOURCE: SAMPLE QUESTION OF BD MATH OLYMPIAD.
In an acute-angled triangle ABC, AD, BE and CF are respectively perpendicular to the opposite side of the three climax point included A, B and C. H is the orthocentre of the orthogonals. What is the value of $ \frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}$?
Liklier figure of the problem:
Thanks in advance for your respective effort.
As you wrote, we have $$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C$$ Now using that $$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1\tag1$$ (the proof for $(1)$ is written at the end of this answer)
we get $$\frac{7}{4R^2}+2\cdot\frac{3}{8R^3}=1,$$ i.e. $$(R+1)(2R-3)(2R+1)=0$$ to have $$\color{red}{R=\frac 32}$$
Let us prove $(1)$.
We have $$\begin{align}\cos 2A+\cos 2B+\cos 2C&=2\cos(A+B)\cos(A-B)+\cos 2C\\\\&=-2\cos C\cos(A-B)+2\cos^2C-1\\\\&=-2\cos C(\cos(A-B)-\cos C)-1\\\\&=-2\cos C\ (\cos(A-B)+\cos(A+B))-1\\\\&=-4\cos A\cos B\cos C-1\end{align}$$ from which we have $$(2\cos^2A-1)+(2\cos^2B-1)+(2\cos^2C-1)=-4\cos A\cos B\cos C-1$$
$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$
$\frac{BC}{AB}=\frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$
$\frac{NM'}{PN}=\sqrt3$
$\frac{BC}{AB}=\sqrt3$
$BC=AB\sqrt3=5\sqrt3$
$a=5,b=3,\color{blue}{a+b=8}$
Best Answer
Let $[PQR]$ denote the area of $\triangle PQR$
Consider the figure . We have :- $$ \frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}=\frac{AD-DH}{AD}+\frac{BE-HE}{BH}+\frac{CF-HF}{CF}$$ $$=3-(\frac{DH}{AD}+\frac{HE}{BH}+\frac{HF}{CF}) $$
Now , $$\frac{DH}{AD}+\frac{HE}{BH}+\frac{HF}{CF} = \frac{\frac{1}{2}BC\cdot DH}{\frac{1}{2}BC \cdot AD}+\frac{\frac{1}{2}AC\cdot HE}{\frac{1}{2}AC \cdot BH}+\frac{\frac{1}{2}AB \cdot HF}{\frac{1}{2}AB \cdot CF} $$ $$= \frac{[BHC]+[CHA]+[HAB]}{[ABC]}=1$$
Therefore , the answer to your question equals $3-1=\boxed2$