Assume a Convex quadrilateral $ABCD$ a,such $|AC|=1,|BC|=3\sqrt{2}$,and $|AD|=|BD|=\frac{\sqrt{2}}{2}|AB|$, find the maximum of the value $|AB|+\sqrt{2}|CD|$

The Book only have answer: $10$.

I wanted to solve this by using law of cosines in triangles $ABC$and $BDC$ to determine $AB$ and $CD$,

Assume that $\angle ACB=x$,then $AB=\sqrt{1+18-6\sqrt{2}\cos{x}}=\sqrt{19-6\sqrt{2}\cos{x}}$,But I can't determine $CD$ .

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    $\begingroup$ So what is your question? $\endgroup$ – Hetebrij Sep 14 '15 at 14:02

Let $\angle ADC=x$, then $\angle BDC=90^\circ-x$. Apply the law of cosine to $\triangle ADC$ we have $$2AD\cdot CD\cos x=AD^2+CD^2-1\tag{1}$$ and to $\triangle BCD$ we have $$2BD\cdot CD \cos (90^\circ-x) = BD^2+CD^2-18\tag{2}.$$ Square both (1) and (2) and add them up we get $$4AD^2\cdot CD^2 = (AD^2+CD^2-1)^2+(AD^2+CD^2-18)^2.$$ so $$0=2AD^4+2CD^4-26(AD^2+CD^2)+145\ge (AD^2+CD^2)^2-38(AD^2+CD^2)+(1+18^2).$$ It follows that $$AD^2+CD^2\le 19+\sqrt{19^2-1-18^2}=19+6=25.$$ Thus $$(AD+CD)^2\le 2(AD^2+CD^2)=50$$ or equivalently $$\frac{AB+\sqrt{2}CD}{\sqrt2}\le \sqrt{2(AD^2+CD^2)}\le \sqrt{50}.$$ So $$AB+\sqrt{2}CD\le 10.$$ The equality happens when $AD=CD=BD$, equivalently $\angle ACB=135^\circ$.

  • $\begingroup$ Last number in (2) should be $-18$. $\endgroup$ – Aretino Sep 14 '15 at 16:00
  • $\begingroup$ Thanks @Aretino , I was wondering why I got the wrong answer :-) $\endgroup$ – Quang Hoang Sep 14 '15 at 16:01
  • $\begingroup$ This still makes the book's answer of $10$ misleading, since $2\sqrt{13+8\sqrt3} = 10.36\ldots$ $\endgroup$ – Théophile Sep 14 '15 at 16:04
  • $\begingroup$ @Théophile In fact the book is correct, see my edit above. $\endgroup$ – Quang Hoang Sep 14 '15 at 16:05
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    $\begingroup$ @QuangHoang Then we can all breathe a sigh of relief. :) $\endgroup$ – Théophile Sep 14 '15 at 16:06

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