The problem is from geometry.

Question

Let $Vambda$ be a positive real number.

Consider a semicircle of center O and diameter

AB. Choose points C and D (C is between D and

B) on the semicircle and let \angle AOD=2\alpha

and \angle BOC=2\beta.

The point X on the line CD is such that XD/

XC=lambda. Prove that when \alpha and \beta

satisfy \tan\alpha=\tan\beta+\sqrt(3)/2, all lines

through X perpendicular to CD pass through a

fixed point.

Answer 1

Step 1 Extend AD and BC to meet at P. Let m denote the line through O and P. We will first show that m is a fixed line.

Let the three altitudes of ΔPAB meet at H. Also let R denote the foot of the altitude from P.

∠AOD = 2α → ∠ACD = α and ∠BOC = 2β → ∠BDC = β.

Since PDHC is cyclic, we also have ∠HPD = α and ∠HPC = β.

We get tan α - tan β = AR/PR - BR/PR = (AO+OR)/PR - (BO-OR)/PR = 2OR/PR = 2/(slope of m).

Given that tan α - tan β = √3/2 is constant, this implies that the slope of m is constant, and hence, m is a fixed line.

Step 2 Let l denote the line through X perpendicular to CD and point G be the intersection between m and l. We will show that G is a fixed point.

Let the two lines perpendicular to AB at B and A meet line m at E and F, respectively. We will show that DE and CF are both perpendicular to CD.

If the above statement is true, we will have DE, CF, and l all parallel to each other. And given that E,F, and G lie on the same line, we must have EG/FG = DX/CX = λ.

Since E and F are intersection points between the fixed line m and the fixed BE and AF, they are both fixed points. It follows that G must also be a fixed point.

First, we will show that DE ⊥ CD.

Extend AB and DC to meet at point Q.

The cyclic ARCP means ∠BCR = ∠BAD. And the cyclic ABCD means ∠BCQ = ∠BAD. Therefore, BC bisects ∠RCQ.

Also, ∠OCR = ∠OCB-∠BCR = ∠OBC-∠BCQ = ∠OQC. So ΔOCR and ΔOQC are similar, and we have OR/OC = RC/CQ. But BC bisects ∠RCQ, so OR/OC = RC/CQ = RB/BQ.

We also know that PR is parallel to EB, so OR/OC = OR/OB = PR/EB. Hence, RB/BQ = PR/EB. This implies that the right triangles ΔPRB and ΔEBQ are similar.

So, we have PB parallel to EQ and therefore, ∠BEQ = ∠EBP.

Since BE is tangent to the circle, we also have ∠EBP = ∠BDC.

So B,D,E,Q are concyclic and therefore, ∠EDQ = ∠EBQ = 90° as desired.

Next, notice that AF and BE are symmetric to the point O, therefore O is the midpoint of EF.

Also notice that the line through O parallel to DE bisects CD.

Hence, CF must be parallel to DE and perpendicular to CD.

We now have DE, CF, and l parallel to each other and conclude that G is the desired fixed point.