Let $ABC$ be an equilateral triangle with circumcircle $k$. A circle $q$ touches $k$ from outside in a point $D$, where the point $D$ on $k$ is chosen so that $D$ and $C$ lie on different sides of the line $AB$. We now draw tangent lines from $A,B$ and $C$ to the circle $q$ and denote the lengths of the respective tangent line segments by $a,b$ and $c$. Prove that $a+b=c$.
Problem
Source: Germany 2022, Problem 5
Tags: geometry, geometry proposed, tangent, tangent lines, touching circles
cadaeibf
25.06.2022 15:26
Trivial by Casey's Theorem
Kugelmonster
28.02.2023 15:01
Lets do some Trigonometry Bashing for this problem
Denote $O$, $Q$ the midpoints of $k$, $q$.
Denote $R$ and $r$ the radii of $k$ and $q$.
We want to use cosine lemma to calculate $a^2, b^2, c^2$ in terms of $R$, $r$ and $\alpha := \angle DOC$:
With cosine lemma in triangle $OBQ$, we get
$$ b^2 = BQ^2 - r^2 = (R+r)^2+R^2-2R(R+r)\cdot \cos (\angle BOD) - r^2 = R(R+r)(2-\sqrt 3 \cdot \sin \alpha + \cos \alpha ) $$since
$$ \cos (\angle BOD) = \cos (\alpha - 120) = \cos \alpha \cos (120) - \sin \alpha - \sin (120) = \frac{- \sqrt 3 \cdot \sin \alpha + \cos \alpha}{2}$$and analogeously using triangle $OAQ$,
$$ a^2 = R(R+r)(2+\sqrt 3 \cdot \sin \alpha + \cos \alpha)$$and using triangle $OCQ$,
$$ c^2 = R(R+r)(2-2\cos\alpha) $$For convenience, set $1 = R(R+r)$
We need to prove that $a+b=c \Leftrightarrow 2ab = c^2-a^2-b^2$:
$$ 2ab =^! 2 - 2\cos \alpha - 2 - \sqrt 3 \cdot \sin \alpha - \cos \alpha - 2 + \sqrt 3 \cdot \sin \alpha - \cos \alpha = -2 -4\cos \alpha$$Both sides have the same sign since $120 < \alpha < 240 \Rightarrow -\frac 12 > \cos \alpha \ge -1 $.
Therefore, we can square both sides and need to prove:
$$ a^2\cdot b^2 =^! (-1-2\cos \alpha)^2 $$$$ \left((2+ \cos \alpha ) + \sqrt 3 \cdot \sin \alpha + \cos \alpha \right) \cdot \left((2+ \cos \alpha) - \sqrt 3 \cdot \sin \alpha\right) =^! 1 + 4\cos \alpha + 4\cos ^2 \alpha $$$$ 4 + 4\cos \alpha + \cos ^2 \alpha - 3 \sin ^2 \alpha =^! 1 + 4\cos \alpha + 4\cos ^2 \alpha $$which is true since $\sin ^2 \alpha + \cos ^2 \alpha = 1$.
Ld_minh4354
17.04.2023 06:43
Let the centers of $(ABC)$ and $q$ be $O$ and $O'$, respectively. Furthermore, let $x=O'D/OD$. Let line $DA$ intersects the circle $q$ at point $E$. Then $a^2=AD\times AE$. Triangles $OAD$ and $O'ED$ are similar by SAS. Thus, $$\frac{DE}{AD}=\frac{O'D}{OD}=x$$$$\Rightarrow a^2 = AD\times AE = AD\times (AD+DE) = AD\times (AD+AD\times x) = AD^2(x+1)$$$$\Rightarrow a = AD \sqrt{x+1}$$ Similarly, $$b = BD \sqrt{x+1}$$$$c = CD \sqrt{x+1}$$ And we are done by the well-known lemma that $DA+DB=DC$.