Let $ABCD$ be a convex quadrilateral such that $\angle ABC = \angle ADC = 135^{\circ}$ and \[AC^2\cdot BD^2 = 2\cdot AB\cdot BC\cdot CD\cdot DA.\]Prove that the diagonals of the quadrilateral $ABCD$ are perpendicular.
Problem
Source: USA TST (42nd IMO-2001)
Tags: geometry, rhombus, inequalities, geometry solved
17.03.2004 20:11
I can't belive I see such an easy problem remain unsolved. 2AB*BC*CD*DA is 4 times the square of the area of ABCD. The conclusion easily follows.
18.03.2004 02:58
I didn't get it. Why did you say that 2AB*BC*CD*DA is 4 times the square of the area of ABCD?
18.03.2004 04:48
A logical flaw...? Assuming the convex quadrilateral is a rhombus then yeah 2*product of sides = 4 times square of area of ABCD. But I think you have to prove that it is a rhombus with the given information.
18.03.2004 11:34
I made a mistake. Not the first, no the last. Hope I don't make any more like this one.
18.03.2004 14:26
Hope there is no mistake this time. Let S be the area of the quadrilateral. Let S1 be the area of ABC and S2 the area of ADC. We have (AC*BD/2)^2 \geq S^2=(S1+S2)^2 \geq 4S1S2=AB*BC*AD*DC/2. The equality takes place iff S1=S2 and AC and BD are perpendicular. Still don't know why nobody solved it . I'll kill you if I made another mistake in this proof.
18.03.2004 17:31
Wow! That's short! I wanted to post my proof, but it's sort of pointless now.. . Anyway, what I did was show that it's true if BD perpendicular to AC, and then prove that it can't be true if we keep A, C, D fixed and give B a "nudge" on the circle ABD.
19.03.2004 18:14
amfulger wrote: Let S be the area of the quadrilateral. Let S1 be the area of ABC and S2 the area of ADC. We have (AC*BD/2)^2 \geq S^2=(S1+S2)^2 \geq 4S1S2=AB*BC*AD*DC/2. The equality takes place iff S1=S2 and AC and BD are perpendicular. Ah, now it's clear: One of the usual problems where some inequalities imply an equality. Darij
20.03.2004 01:29
darij grinberg wrote: Ah, now it's clear: One of the usual problems where some inequalities imply an equality. Darij The inequality did not imply equality, the equality was the special case amfulger was looking for.
26.12.2019 09:38
Seriously? \begin{align*} (AD \cdot BC)^2 &\ge (2[ABCD])^2 = 4([ABC]+[ADC])^2 \\ &\ge 4\left(2\sqrt{[ABC] \cdot [ADC]}\right)^2 = 16 \cdot \left(\dfrac{1}{2} \cdot \sin 135^\circ \right)^2 (AB \cdot BC)(CD \cdot DA) \\ &= 2 \cdot AB \cdot BC \cdot CD \cdot DA.\end{align*}Equality only if $AD \cdot BC = 2[ABCD]$ i.e. $\overline{AD} \perp \overline{BC}$. $\blacksquare$
05.06.2020 22:18
nice joke If the angle made by $\overline{AC}$ and $\overline{BD}$ is $\theta$ and $K=2[ABCD]$, then by AM-GM $$AC^2 \cdot BD^2 = \left( \frac{K}{\sin(\theta)} \right)^2 \geq \left( \cos(135^\circ)(AB \cdot BC+CD \cdot DA ) \right)^2 \geq 2 \cdot AB \cdot BC \cdot CD \cdot DA$$with equality only if $\theta=90^\circ$.
15.01.2021 18:56
Let $K,L$ be the feet from $B,D$ to line $AC$. Observe that \[\frac{1}{16}AC^2\cdot BD^2=[ABC]\cdot [ADC]=\frac{1}{4}AC^2\cdot BK\cdot DL.\]Then note the inequality $4BK\cdot DL\le (BK+DL)^2\le BD^2$ so $B$ and $D$ reflect to each other over line $AC$, done.
15.01.2021 20:26
$AC^2.BD^2=4(AB.BC sin 135)(CD.DA.sin135) \implies AC^2.BD^2=16.S(\triangle ABC).S(\triangle ACD) \implies $ if we draw altitudes DD' and BB' of this triangles then problem equivalent to: $BD^2=2DD'.BB' $ let diagnals meet at $R$ then$ BD^2=B'R^2+D'R^2+BB'^2+DD'^2 \geq 2DD'.BB' \implies RD' ,RB'=0 $ $ \blacksquare$
13.02.2021 16:48
Does anybody have a non-bashing solution ? Mine was through calculating the Areas of $ABC$,$ADC$ and $ABCD$
13.02.2021 19:41
Outline of my solution. Let $X$ be the intersection of diagonals. Using law of sines on $\triangle ABC$, $\triangle ACD$, $\triangle AXD$ and $\triangle AXB$ and using the given equality we obtain that $AD\cdot CD=AB\cdot BC$. Then, we use law of cosines on $\triangle ABC$ and $\triangle ACD$, and we obtain that $AB^2+BC^2=AD^2+CD^2$, squaring both sides here, we have $AB^4+BC^4=AD^4+CD^4$, but notice that this is equivalent to $AB^4+BC^4-2AB\cdot BC=AD^4+CD^4-2AD\cdot CD\Longleftrightarrow (AB^2-BC^2)=(AD^2-CD^2)^2$.
Hence, we have $(AB^2-BC^2)=(AD^2-CD^2)^2\implies AB^2+CD^2=BC^2+AD^2$, meaning that $AC\perp BD$.
17.06.2021 01:28
Let $\theta$ be the angle between $\overline{BD}$ and $\overline{AC}$. Consider the following chain of inequalities: \begin{align*} \frac{1}{4} AC^2 \cdot BD^2 &\geq \left( \frac{1}{2} AC \cdot BD \cdot \sin \theta \right)^2 \\ &= [ABCD]^2 \\ &= \left(\frac 1 2 \sin 135^\circ \right)^2 \cdot (AD \cdot CD + AB \cdot BC)^2 \\ &\geq \frac 1 8 \cdot 4 \cdot AB \cdot BC \cdot CD \cdot DA. \end{align*}In order for equality to hold, we need $\sin \theta = 1$, as desired.
07.02.2023 23:49
We claim that \[AC^2 \cdot BD^2 \geq 2\cdot AB \cdot BC \cdot CD\cdot DA\]with equality only when $AC\perp BD$. By interpreting both sides as areas, the problem becomes a lot more clear. We have the following identities: \begin{align} [ABC] &= \frac{AB\cdot BC \sin 135}{2} = \frac{\sqrt{2}}{4} \cdot AB\cdot BC\\ [ADC] &= \frac{ AD\cdot DC \sin 135}{2}=\frac{\sqrt{2}}{4} \cdot AD\cdot DC \\ [ABCD] &= \frac{AC\cdot BD \sin \theta}{2} \end{align}where $\theta$ is the angle between $AC$ and $BD$. Thus, \begin{align*} (AC\cdot BD)^2 &\stackrel{(3)}{\geq} (2[ABCD])^2 = 4 ([ABC]+[ADC])^2 \\ &\geq 16 [ABC]\cdot [ADC] = 16\cdot \left(\frac{\sqrt{2}}{4}\right)^2 \cdot AB\cdot BC\cdot AD\cdot DC\\ &= 2\cdot AB\cdot BC\cdot CD\cdot DA, \end{align*}and we have shown the inequality. Equality holds when $\sin \theta = 1$, and $[ABC] = [ADC]$. Luckily, both of these are equivalent to the diagonals being perpendicular, and we're done. $\blacksquare$.
20.02.2024 04:58
Observe the inequality $$AC \cdot BD \geq 2[ABCD] = \frac{AD \cdot CD + AB \cdot BC}{\sqrt 2} \geq \sqrt{2 \cdot AB \cdot BC \cdot CD \cdot DA}$$by AM-GM. Thus equality holds when $AC \cdot BD = 2[ABCD]$ precisely, i.e. $ABCD$ is orthodiagonal.
20.07.2024 07:55
Notice $2AB\cdot BC\cdot CD\cdot DA=16[ABC][ADC].$ Letting $h_B,h_D$ be the distances from $B,D$ to $AC,$ we have $BD^2=4h_Bh_D$ but $4h_Bh_D\le(h_B+h_D)^2\le BD^2,$ the last equality holding only if $BD\perp AC.$