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Geometry Difficulty 6.6 National olympiad Prove it Brazil

A convex quadrilateral is inscribed in a circle of radius 11. Show that its perimeter minus the sum of its two diagonals lies between 00 and 22.

Solution

Let the quadrilateral be ABCDABCD. We have AB+BC>ACAB + BC > AC and CD+DA>ACCD + DA > AC, so the perimeter is bigger than 2⋅AC2 \cdot AC and, similarly, bigger than 2⋅BD2 \cdot BD. Hence it's bigger than AC+BDAC + BD, and the perimeter minus the sum of the diagonals is bigger than zero. Notice that this inequality is sharp: just consider a rectangle with two sides next to zero.

Figure 1

Now, α+β+γ+δ=180∘\alpha + \beta + \gamma + \delta = 180^\circ and, since the radius of the circle is 11, AB=2sin⁡γAB = 2 \sin \gamma, BC=2sin⁡βBC = 2 \sin \beta, CD=2sin⁡αCD = 2 \sin \alpha, AD=2sin⁡δAD = 2 \sin \delta, AC=2sin⁡(β+γ)=2sin⁡(α+δ)=sin⁡(β+γ)+sin⁡(α+δ)AC = 2 \sin(\beta + \gamma) = 2 \sin(\alpha + \delta) = \sin(\beta + \gamma) + \sin(\alpha + \delta) and BD=2sin⁡(α+β)=2sin⁡(γ+δ)=sin⁡(α+β)+sin⁡(γ+δ)BD = 2 \sin(\alpha + \beta) = 2 \sin(\gamma + \delta) = \sin(\alpha + \beta) + \sin(\gamma + \delta), so the difference required is
2(sin⁡α+sin⁡β+sin⁡γ+sin⁡δ)−sin⁡(α+δ)−sin⁡(β+γ)−sin⁡(α+β)−sin⁡(γ+δ) 2(\sin \alpha + \sin \beta + \sin \gamma + \sin \delta) - \sin(\alpha + \delta) - \sin(\beta + \gamma) - \sin(\alpha + \beta) - \sin(\gamma + \delta)
Notice that sin⁡(α+γ)\sin(\alpha + \gamma) and sin⁡(β+δ)\sin(\beta + \delta) are “missing”. So we will prove a stronger result, that is,
E=2(sin⁡α+sin⁡β+sin⁡γ+sin⁡δ)−sin⁡(α+δ)−sin⁡(β+γ)−sin⁡(α+β)−sin⁡(γ+δ)−sin⁡(α+γ)−sin⁡(β+δ)<0  ⟹  2(sin⁡α+sin⁡β+sin⁡γ+sin⁡δ)−sin⁡(α+δ)−sin⁡(β+γ)−sin⁡(α+β)−sin⁡(γ+δ)<sin⁡(α+γ)+sin⁡(β+δ)≤2 \begin{aligned} E &= 2(\sin \alpha + \sin \beta + \sin \gamma + \sin \delta) \\ &\quad - \sin(\alpha + \delta) - \sin(\beta + \gamma) - \sin(\alpha + \beta) - \sin(\gamma + \delta) - \sin(\alpha + \gamma) - \sin(\beta + \delta) < 0 \\ \implies \quad & 2(\sin \alpha + \sin \beta + \sin \gamma + \sin \delta) - \sin(\alpha + \delta) - \sin(\beta + \gamma) - \sin(\alpha + \beta) - \sin(\gamma + \delta) \\ &< \sin(\alpha + \gamma) + \sin(\beta + \delta) \le 2 \end{aligned}
First notice that
sin⁡(α+β)+sin⁡(α+γ)=2sin⁡(α+β+γ2)cos⁡(β−γ2)=2cos⁡(α−δ2)cos⁡(β−γ2) \begin{aligned} \sin(\alpha + \beta) + \sin(\alpha + \gamma) &= 2 \sin \left( \alpha + \frac{\beta + \gamma}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) \\ &= 2 \cos \left( \frac{\alpha - \delta}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) \end{aligned}
so
sin⁡β+sin⁡γ−sin⁡(α+β)−sin⁡(α+γ)=2sin⁡(β+γ2)cos⁡(β−γ2)−2cos⁡(α−δ2)cos⁡(β−γ2)=2cos⁡(β−γ2)(cos⁡(α+δ2)−cos⁡(α−δ2))=−4cos⁡(β−γ2)sin⁡α2sin⁡δ2 \begin{aligned} & \sin \beta + \sin \gamma - \sin(\alpha + \beta) - \sin(\alpha + \gamma) \\ &= 2 \sin \left( \frac{\beta + \gamma}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) - 2 \cos \left( \frac{\alpha - \delta}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) \\ &= 2 \cos \left( \frac{\beta - \gamma}{2} \right) \left( \cos \left( \frac{\alpha + \delta}{2} \right) - \cos \left( \frac{\alpha - \delta}{2} \right) \right) \\ &= -4 \cos \left( \frac{\beta - \gamma}{2} \right) \sin \frac{\alpha}{2} \sin \frac{\delta}{2} \end{aligned}
Summing three analogous equations, we obtain
2(sin⁡α+sin⁡β+sin⁡γ)−sin⁡(α+δ)−sin⁡(β+γ)−sin⁡(α+β)−sin⁡(γ+δ)−sin⁡(α+γ)−sin⁡(β+δ)=−4sin⁡δ2(cos⁡(β−γ2)sin⁡α2+cos⁡(α−γ2)sin⁡β2+cos⁡(β−α2)sin⁡γ2)=−4sin⁡δ2(sin⁡(α+β−γ2)+sin⁡(α−β+γ2)+sin⁡(−α+β+γ2)) \begin{aligned} & 2(\sin \alpha + \sin \beta + \sin \gamma) \\ &\quad - \sin(\alpha + \delta) - \sin(\beta + \gamma) - \sin(\alpha + \beta) - \sin(\gamma + \delta) - \sin(\alpha + \gamma) - \sin(\beta + \delta) \\ &= -4 \sin \frac{\delta}{2} \left( \cos \left( \frac{\beta - \gamma}{2} \right) \sin \frac{\alpha}{2} + \cos \left( \frac{\alpha - \gamma}{2} \right) \sin \frac{\beta}{2} + \cos \left( \frac{\beta - \alpha}{2} \right) \sin \frac{\gamma}{2} \right) \\ &= -4 \sin \frac{\delta}{2} \left( \sin \left( \frac{\alpha + \beta - \gamma}{2} \right) + \sin \left( \frac{\alpha - \beta + \gamma}{2} \right) + \sin \left( \frac{-\alpha + \beta + \gamma}{2} \right) \right) \end{aligned}
Finally,
E=2sin⁡δ−4sin⁡δ2(sin⁡(α+β−γ2)+sin⁡(α−β+γ2)+sin⁡(−α+β+γ2))=−4sin⁡δ2(−cos⁡δ2+sin⁡(α+β−γ2)+sin⁡(α−β+γ2)+sin⁡(−α+β+γ2))=−4sin⁡δ2(−sin⁡(α+β+γ2)+sin⁡(α+β−γ2)+2cos⁡(α−β2)sin⁡γ2)=−8sin⁡δ2(−cos⁡(α+β2)sin⁡γ2+cos⁡(α−β2)sin⁡γ2)=−8sin⁡γ2sin⁡δ2(−cos⁡(α+β2)+cos⁡(α−β2))=−16sin⁡α2sin⁡β2sin⁡γ2sin⁡δ2<0 \begin{align*} E &= 2\sin\delta - 4\sin\frac{\delta}{2}\left(\sin\left(\frac{\alpha+\beta-\gamma}{2}\right) + \sin\left(\frac{\alpha-\beta+\gamma}{2}\right) + \sin\left(\frac{-\alpha+\beta+\gamma}{2}\right)\right) \\ &= -4\sin\frac{\delta}{2}\left(-\cos\frac{\delta}{2} + \sin\left(\frac{\alpha+\beta-\gamma}{2}\right) + \sin\left(\frac{\alpha-\beta+\gamma}{2}\right) + \sin\left(\frac{-\alpha+\beta+\gamma}{2}\right)\right) \\ &= -4\sin\frac{\delta}{2}\left(-\sin\left(\frac{\alpha+\beta+\gamma}{2}\right) + \sin\left(\frac{\alpha+\beta-\gamma}{2}\right) + 2\cos\left(\frac{\alpha-\beta}{2}\right)\sin\frac{\gamma}{2}\right) \\ &= -8\sin\frac{\delta}{2}\left(-\cos\left(\frac{\alpha+\beta}{2}\right)\sin\frac{\gamma}{2} + \cos\left(\frac{\alpha-\beta}{2}\right)\sin\frac{\gamma}{2}\right) \\ &= -8\sin\frac{\gamma}{2}\sin\frac{\delta}{2}\left(-\cos\left(\frac{\alpha+\beta}{2}\right) + \cos\left(\frac{\alpha-\beta}{2}\right)\right) \\ &= -16\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\sin\frac{\delta}{2} < 0 \end{align*}
because 0<α2,β2,γ2,δ2<90∘0 < \frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2}, \frac{\delta}{2} < 90^\circ.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.