IMO ShortList 1998, geometry problem 2

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IMO ShortList 1998, geometry problem 2

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Source: IMO ShortList 1998, geometry problem 2

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orl

#1 h Oct 22, 2004, 2:22 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$ , respectively, such that $AE:EB=CF:FD$ . Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$ . Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$ .

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orl

#2 h Oct 22, 2004, 2:22 PM • 2 Y

Y by Adventure10, Mango247

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

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#3 h Oct 22, 2004, 7:30 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

At first we note an important result:

Lemma 1. If A and B are two points in the plane, and X and Y are two other points, and Z is the point on the line XY which divides the segment XY in the ratio XZ : ZY = p : q (with directed segments), then

$\left[ AZB\right] =\frac{q\cdot \left[ AXB\right] +p\cdot \left[ AYB\right] }{q+p}$ ,

where $\left[P_1P_2P_3\right]$ denotes the directed area of a triangle $P_1P_2P_3$ . ("Directed" means that the area is taken positive if the triangle is orientated counterclockwise, and negative if it is orientated clockwise.)

Proof. Let x, y, z be the directed distances from the points X, Y, Z to the line AB. ("Directed" means that e. g. the distance x is positive if the triangle AXB is orientated counterclockwise, and negative if it is orientated clockwise.) Then, since the area of a triangle equals $\frac12 \cdot$ sidelength $\cdot$ corresponding altitude, we have

$\left[ AXB\right] =\frac{1}{2}\cdot AB\cdot x$ ;
$\left[ AYB\right] =\frac{1}{2}\cdot AB\cdot y$ ;
$\left[ AZB\right] =\frac{1}{2}\cdot AB\cdot z$ .

Thus, the areas [AXB], [AYB], [AZB] are proportional to the distances x, y, z, respectively. Hence, instead of proving $\left[ AZB\right] =\frac{q\cdot \left[ AXB\right] +p\cdot \left[ AYB\right] }{q+p}$ , it is enough to show $z =\frac{q\cdot x +p\cdot y}{q+p}$ . But this is clear if you consider the line AB as the x-axis of an Cartesian coordinate system (then x, y, z are the x-coordinates of the points X, Y, Z, respectively). Lemma 1 is proven.

Another important and well-known lemma:

Lemma 2. If ABC is a triangle with sidelengths a = BC, b = CA, c = AB and circumradius R, then the area of triangle ABC equals $\frac{abc}{4R}$ .

Now, in our cyclic quadrilateral ABCD, let a = AB, b = BC, c = CD, d = DA, e = AC, f = BD. Let R be the circumradius of the quadrilateral ABCD. And finally, call p : q = AE : EB = CF : FD.

We will apply Lemma 1 but work with non-directed areas. In other words, in the following, the notation $\left[P_1P_2P_3\right]$ will mean the non-directed area of a triangle $P_1P_2P_3$ . Of course, one can apply Lemma 1 for non-directed areas only if the points X, Y, Z all lie on one and the same side of the line AB; but this will be always guarranteed in the following.

Applied to triangle ABD, whose sidelengths are AB = a, BD = f and DA = d, and whose circumradius is R, Lemma 2 yields

$\left[ ABD\right] =\frac{afd}{4R}$ .

Since the point E lies on the line AB and we have AE : EB = p : q, Lemma 1 gives

$\left[ AED\right] =\frac{q\cdot \left[ AAD\right] +p\cdot \left[ ABD\right] }{q+p}=\frac{q\cdot 0+p\cdot \left[ ABD\right] }{q+p}=\frac{p\cdot \left[ ABD \right] }{q+p}$
$=\frac{p}{q+p}\cdot \left[ ABD\right] =\frac{p}{q+p}\cdot \frac{afd}{4R}$ .

Similarly,

$\left[ AFD\right] =\frac{q}{q+p}\cdot \frac{ced}{4R}$ .

Since the point P lies on the line EF and we have PE : PF = AB : CD = a : c, Lemma 1 yields

$\left[ APD\right] =\frac{c\cdot \left[ AED\right] +a\cdot \left[ AFD\right] }{c+a}=\frac{c\cdot \frac{p}{q+p}\cdot \frac{afd}{4R}+a\cdot \frac{q}{q+p} \cdot \frac{ced}{4R}}{c+a}=\frac{\left( pf+qe\right) acd}{4R\left( p+q\right) \left( c+a\right) }$ .

Similarly,

$\left[ BPC\right] =\frac{\left( pf+qe\right) acb}{4R\left( p+q\right) \left( c+a\right) }$ .

Thus,

$\frac{\left[ APD\right] }{\left[ BPC\right] }=\frac{d}{b}$ ,

what is independent from the choice of E and F indeed. Proof complete.

Darij

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#4 h Jun 28, 2006, 1:15 AM • 6 Y

Y by Sx763_, Adventure10, Mango247, Stuffybear, and 2 other users

If $BC$ and $DA$ are parallel, then it's trivial, because then $ABCD$ would be an isosceles trapezoid, so in fact, we would have $AE=CF$ and then it's easy to see that $d(P,BC)=d(P,DA)$ .

So let's now assume that $BC$ and $DA$ intersect at a point $G$ .
We have $AB: CD=AE: CF=EP: FP$ . But since $ABCD$ is cyclic, the triangles $GAB$ and $GCD$ are similar with ratio $AB: CD$ . But since $AE: CF$ is equal to the similtude ratio, the point $E$ in $GAB$ corresponds to the point $F$ in $GCD$ . Thus, the triangles $GFC$ and $GEA$ are similar with ratio $AB: CD$ so we have $GE: GF=AB: CD$ and $\angle DEF = \angle CGE$ Thus, $GE: GF=EP: FP$ and hence, $GP$ is the angle bisector of $\angle AGB$ from which we conclude $d(P,AD)=d(P,BC)$ , yielding that the areas in question have a same attitude, thus their ratio is equal to the ratio of the corresponding sides, i.e. $AD: BC$

$\Box$

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NewAlbionAcademy

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NewAlbionAcademy

#5 h May 27, 2013, 2:56 AM • 3 Y

Y by rashah76, Adventure10, Mango247

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#6 h Dec 30, 2019, 9:26 PM • 2 Y

Y by teomihai, Adventure10

1998 G2 wrote:

Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$ , respectively, such that $AE:EB=CF:FD$ . Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$ . Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$ .

Tutorial for linearity

Animate $E$ linearly on side $AB$ . Then $E \mapsto F$ is linear and since $P$ divides $EF$ in a fixed ratio, $P$ moves linearly. Thus, it suffices to show that the locus of $P$ passes through $Q=AD \cap BC$ , to conclude that $\frac{[APD]}{[BPC]}$ is fixed.

Let $X$ and $Y$ be points on $AC$ and $BD$ such that $\frac{AX}{XC}=\frac{BY}{YD}=\frac{AB}{CD}$ . When $E=A$ we have $P=X$ and when $E=B$ we have $P=Y$ . Thus, $\overline{XY}$ is the locus of $P$ . We want to show that $Q, X, Y$ are collinear. Notice that $\frac{QA}{QC}=\frac{QB}{QD}=\frac{AB}{CD}$ hence $X, Y$ lie on the internal bisector of angle $BQD$ , proving the result. $\blacksquare$

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awesomeming327.

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awesomeming327.

#7 h Jul 13, 2022, 3:01 AM

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Diagram

Let $AE=x,BE=ax,CF=y,DF=ay.$ It follows that $AB:CD=x:y$ so $PE:PF=x:y.$ Define $d(X,YZ)$ to the distance between point $X$ to line $YZ$ for any $X,Y,Z.$

We have $d(E,AD)=x\sin(\angle A),d(F,AD)=ay\sin (\angle C).$ Since $P$ is $\frac{x}{x+y}$ of the way from $E$ to $F$ we have $d(P,AD)=\frac{yd(E,AD)+xd(F,AD)}{x+y}=\frac{xy}{x+y}\left(\sin(\angle A)+a\sin\angle C\right)$ Similarly, we have $d(P,BC)=\frac{xy}{x+y}\left(\sin(\angle C)+a\sin\angle D\right)$ and since $ABCD$ is cyclic, we see that these two are actually equal. Thus, we have $\frac{[APD]}{[BPC]}=\frac{AD\cdot d(P,AD)}{BC\cdot d(P,BC)}=\frac{AD}{BC}$ which is not dependant on the choice of $a$ as desired.

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#8 h Jul 25, 2022, 9:51 PM

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G2?

Let $\frac{AE}{AB}=\frac{CF}{CD}=k$ . Note that
$[APD]=[AED]\cdot \frac{CD}{AB+CD}+[AFD]\cdot \frac{AB}{AB+CD}$ which simplifies to
$[APD]=k\cdot [ABD]\cdot \frac{CD}{AB+CD}+(1-k)\cdot [ACD]\cdot \frac{AB}{AB+CD}.$ Repeating similar calculations, we find that
$\frac{[APD]}{[BPC]}=\frac{k\cdot [ABD]\cdot CD+(1-k)\cdot [ACD]\cdot AB}{(1-k)\cdot [ABC]\cdot CD+k\cdot [BDC]\cdot AB}.$ It is then enough to show that
$\frac{[ABD]\cdot CD}{[BDC]\cdot AB}=\frac{[ACD]\cdot AB}{[ABC]\cdot CD}.$ But now we have $\frac{[ABD]\cdot CD}{[BDC]\cdot AB}=\frac{AB\cdot CD\cdot AD}{AB\cdot CD\cdot BC}=\frac{AD}{BC}$ . The RHS is also equal to $\frac{AD}{BC}$ , so we are done. $\blacksquare$

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#9 h Mar 28, 2023, 8:59 PM

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MMP >

Note that as $E$ moves along $AB$ , $F$ moves linearly along $BC$ . Furthemore, $P$ is a weighted mean of $E$ and $F$ , so it also moves linearly as $E$ moves.

Let $P_1$ be the point on $AC$ such that $\frac{AP_1}{P_1C}=\frac{AB}{CD}.$ Similarly, let $P_2$ be the point on $BD$ such that $\frac{BP_2}{P_2D}=\frac{AB}{CD}.$ Finally, let $P_3$ be the intersection of $AD$ and $BC$ (possibly a point at infinity).

If $E=A,$ then $F=C$ and $P=P_1$ . Similarly, if $E=B$ , then $F=D$ and $P=P_2$ . Therefore, since $P$ moves linearly, $P$ always lies on line $P_1P_2$ . We want to show that the ratio of the heights from $P$ to $AD$ and $BC$ is constant. Thus, it suffices to show that $P_1,P_2,P_3$ are collinear.

Note that from similar triangles $\triangle P_3AB$ and $\triangle P_3CD,$ $\frac{AB}{CD}=\frac{P_3A}{P_3C}.$ Hence, $\frac{AP_1}{P_1C}=\frac{AB}{CD}=\frac{P_3A}{P_3C},$ so $P_1$ lies on the bisector of $\angle AP_3C$ . Similarly, $P_2$ lies on the bisector of $\angle BP_3D$ , so $P_1,P_2,P_3$ are collinear and we are done.

This post has been edited 1 time. Last edited by john0512, Mar 28, 2023, 9:00 PM

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ezpotd

#10 h Oct 13, 2024, 12:34 AM

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Let $d(X,l)$ denote the distance from point $X$ to line $l$ . We desire to show that $d(P, AD) : d(P, BC)$ is constant regardless of $E,F$ . We calculate $d(P,AD) =\frac{CD \cdot d(E, AD)+ AB \cdot d(F,AD)}{AB+ CD} = \frac{x \cdot CD \cdot AB \sin A + (1 - x) \cdot AB \cdot CD \sin B}{AB + CD} = \frac{AB \cdot CD (x \sin A + (1 - x) \sin B)}{AB + CD}$ , $d(P, BC) = \frac{CD \cdot d(E, BC) + AB \cdot d(F,BC)}{AB + CD} = \frac{(1 - x) \cdot CD \cdot AB \sin B + x \cdot AB \cdot \sin A}{AB + CD} = \frac{AB \cdot CD (x \sin A + (1 -x) \sin B)}{AB + CD}$ as desired.

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