#9) Trig Bash
by ayan.nmath, Feb 12, 2018, 4:28 PM
Problem. In 
, let 
be a point on segment 
. Let 
be feet of perpendiculars from onto 
and 
respectively. Suppose 
, 
and 
are concurrent. Let 
be diametrically opposite points of in the circumcircles of triangles 
and 
respectively. Prove that 
, 
and the altitude from 
onto are concurrent.
(Supercali)
, let 
be a point on segment 
. Let 
be feet of perpendiculars from onto 
and 
respectively. Suppose 
, 
and 
are concurrent. Let 
be diametrically opposite points of in the circumcircles of triangles 
and 
respectively. Prove that 
, 
and the altitude from 
onto are concurrent.(Supercali)
be the midpoint of 
where 
is the foot of 
A
Similarly,![\[\frac{CF}{FA}=\frac{CD\cos C}{b-CD\cos C}\]](http://latex.artofproblemsolving.com/f/f/d/ffd59570fcac16d431a85c6413034a124653d68e.png)
Applying Ceva's theorem,
We also have ![\[CD+BD=a.\]](http://latex.artofproblemsolving.com/c/e/6/ce65ca77138d77458a58e29765370fd298252086.png)
Therefore,![\[BD=\dfrac{a\cos B\cos C-b\cos B+c\cos C}{2\cos B\cos C}\qquad (\spadesuit).\]](http://latex.artofproblemsolving.com/a/a/e/aae508a18636cbd55880feadcb439858d5040fee.png)
it follows that![\[\angle ABY=90^{\circ}-\angle B,\qquad \angle BAY=\angle EDA=90^{\circ}-\angle BAD\]](http://latex.artofproblemsolving.com/2/f/1/2f121dfa27a591ba29cc46bcb54ee8a86ea1c51f.png)
Using Sine Law in ![\[\frac{c}{\sin\angle BYA}=\frac{YB}{\cos\angle BAD}\]](http://latex.artofproblemsolving.com/8/1/1/811abf1229b8e89caa7de668c7f8662477aa7081.png)
Therefore,
Now, note that,![\[\cos B=\frac{c^2+BD^2-AD^2}{2c\cdot BD}\implies AD^2-BD^2=c^2-2c\cdot BD\cos B\]](http://latex.artofproblemsolving.com/8/6/3/86353bd7471d4cf8a2af338a78e62a25d302fda9.png)
Substituting the above in 
Consider,
W.L.O.G. assume 
Then 
Putting the above in the denominator of 
Thus, ![\[YB=\frac{\Delta}{b\cos C}.\]](http://latex.artofproblemsolving.com/6/b/5/6b5701a3a03ef421221c15e6935e2165e57e1c77.png)
Let 
Then as 
,![\[PB=\frac{a\cdot\tfrac{bc}{4R}}{CA'}=\frac{\Delta}{CA'}=\frac{\Delta}{b\cos C}\]](http://latex.artofproblemsolving.com/0/f/1/0f12c7804ab3b5dadfca6dfc5b6eec66ce13f60d.png)
Thus we obtain 
Hence 
And we are done. 
![[asy]import olympiad;
unitsize(1.5cm);
pair A=(0.8,2), B=(0,0), C=(3, 0), D=relpoint(B--C, 0.8), F = foot(D, A, B), E = foot(D, A, C), O_1 = circumcenter(A,B,D), O_2 = circumcenter(A,D,C), Y = rotate(180,O_1)*D, X=rotate(180,O_2)*D, Z, K = foot(A,B,C);
draw(A--B--C--cycle);
draw(D--E, dashed+0.8blue); draw(D--F,dashed+0.8blue);
draw(rightanglemark(D,E,C,4),0.8blue); draw(rightanglemark(D,F,B,4),0.8blue);
path U = circumcircle(A,B,D);
path V = circumcircle(A,C,D);
draw(U, linetype("3 4")+red);draw(V,linetype("3 4")+red);
draw(B--E, linetype("2 4 6 4")+0.7green); draw(C--F, linetype("2 4 6 4")+0.7green); draw(A--D, linetype("2 4 6 4")+0.7green);
dot(intersectionpoint(B--E, C--F),0.7green);
Z = intersectionpoint(B--X,C--Y);
draw(B--X, orange);draw(C--Y, orange);draw(A--K, orange);
draw(rightanglemark(A,K,C, 4), orange);
dot(A);dot(B);dot(C);dot(D,0.8red);dot(E,0.8blue);dot(F,0.8blue);dot(X,0.8red);dot(Y,0.8red);dot(Z, orange);dot(K,orange);
label("$A$", A, dir(120));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, dir(-75), 0.8red);
label("$E$", E, dir(45),0.8blue);
label("$F$", F, dir(135),0.8blue);
label("$X$", X, dir(45), 0.8red);
label("$Y$", Y, dir(135), 0.8red);
label("$A'$", K, S, orange);
label("$M$", Z, dir(20), orange);[/asy]](http://latex.artofproblemsolving.com/5/b/5/5b5c4976812577ada4205dacc53e072392d87212.png)
This post has been edited 2 times. Last edited by pro_4_ever, Mar 1, 2018, 10:45 AM
,
,
![\[\text{Similarly, } \frac{AF}{FB} = \frac{\tan B}{\tan\beta} \quad (***)\]](http://latex.artofproblemsolving.com/9/0/1/901618ab5b4e459136b39ec774a5a213ed099a64.png)
By Ceva's Theoerem and 
,
and 
,
Note that 
and 
.
where 
.
.
![\[\text{Similarly}, Z_2K = \frac{m}{\tan\beta} \cdot \frac{CK}{a}.\]](http://latex.artofproblemsolving.com/a/d/7/ad7650f2f349488919a098d835e2d196c5eea505.png)
Using the above results we have,
Since, 
and 
![\[\implies AK, BX, CY \text{ are concurrent}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/0/6/3/06354f20bcf37515ec5b521736fe7d7da5019ecd.png)
![[asy]import olympiad;
unitsize(1.5cm);
pair A=(0.8,2), B=(0,0), C=(3, 0), D=relpoint(B--C, 0.8), F = foot(D, A, B), E = foot(D, A, C), O_1 = circumcenter(A,B,D), O_2 = circumcenter(A,D,C), Y = rotate(180,O_1)*D, X=rotate(180,O_2)*D, Z, K = foot(A,B,C);
draw(A--B--C--cycle);
draw(D--E, dashed+0.8blue); draw(D--F,dashed+0.8blue);
draw(rightanglemark(D,E,C,4),0.8blue); draw(rightanglemark(D,F,B,4),0.8blue);
path U = circumcircle(A,B,D);
path V = circumcircle(A,C,D);
draw(U, linetype("3 4")+red);draw(V,linetype("3 4")+red);
draw(B--E, linetype("2 4 6 4")+0.7green); draw(C--F, linetype("2 4 6 4")+0.7green); draw(A--D, linetype("2 4 6 4")+0.7green);
dot(intersectionpoint(B--E, C--F),0.7green);
Z = intersectionpoint(B--X,C--Y);
draw(B--X, orange);draw(C--Y, orange);draw(A--K, orange);
draw(rightanglemark(A,K,C, 4), orange);
dot(A);dot(B);dot(C);dot(D,0.8red);dot(E,0.8blue);dot(F,0.8blue);dot(X,0.8red);dot(Y,0.8red);dot(Z, orange);dot(K,orange);
label("$A$", A, dir(120));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, dir(-75), 0.8red);
label("$E$", E, dir(45),0.8blue);
label("$F$", F, dir(135),0.8blue);
label("$X$", X, dir(45), 0.8red);
label("$Y$", Y, dir(135), 0.8red);
label("$K$", K, S, orange);
label("$Z$", Z, dir(20), orange);[/asy]](http://latex.artofproblemsolving.com/7/5/a/75a4ca8d0754a7c3521138fd5c945c4c04d69d9a.png)
March 2018
February 2018