1998 IMO Problems/Problem 1

Problem

In the convex quadrilateral , the diagonals and are perpendicular and the opposite sides and are not parallel. Suppose that the point , where the perpendicular bisectors of and meet, is inside . Prove that is a cyclic quadrilateral if and only if the triangles and have equal areas.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. First, let's prove that if ABCD is cyclic, then the triangles ABP and CDP have equal areas. The circumcenter of ABCD is the intersection of perpendicular bisectors of the sides of the quadrilateral ABCD, so P is the circumcenter. From this, we derive that PC=PB=PA=PD. Since the diagonals AC and BD are perpendicular, the sum of the angles APB and CPD is equal to 180 degrees. Denote the angle APB by a. Then the angle CPD is equal to 180 - a. The area of the triangle APB is equal to AP*BP*sin(angle APB). Let R denote the radius of the circle. Then the area of the triangle APB is equal to R*R*sin(a), and the area of CPD equals to R*R*sin(180 - a) = R*R*sin(a), so the triangles ABP and CDP have equal areas. Now, let's prove it in the other direction. We know that ABP and CDP have equal areas. Let O denote the intersection of the diagonals AC and BD. Let M be the midpoint of AB and N be the midpoint of CD. Draw a segment between M and O, and draw another segment between O and N. We know that ON = CN = ND and OM = MB = AM. Since S(ABP) = S(CPD) (where S denotes the area of the triangles), we have AB*PM = CD*PN. From this, we can conclude that PM/PN = CD/AB = ON/OM. Let the intersection of the lines AB and CD be the point E. Since PM is perpendicular to AB and PN is perpendicular to CD, PMEN is a cyclic quadrilateral, so the angle MPN is equal to 180-(angle AED) = BAD + CDA (where BAD and CDA are angles). The angle MON is equal to BAC + CDB + 90 (since the triangles AMO and OND are isosceles). In triangle AOD, OAD+ODA = 90 (where OAD and ODA are angles), so the angle MON is equal to BAC + CDB + CAD + BDA = BAD + CDA = MPN. Now take a look at triangles MPN and NOM. We know that MP/PN = ON/OM and the angle MPN is equal to MON. This implies that the triangle MPN is similar to NOM. Since the two triangles have a common side, MN, we conclude that the triangles MPN and NOM are equal, so MP = ON = CN = CD/2, and PN = OM = MB = AB/2. From this, we can derive that the triangles MBP and NPD are equal, so the angle MPB = angle PDN. Denote the angle by x. In the same way, we can derive that the triangles MBP, MAP, NPC, and NPD are equal, so the sum of the angles APB and CPD is 180 degrees, and the angle APB is equal to 2x. So the area of ABP is equal to AP*AP*sin(2x) and the area of CPD is equal to CP*CP*sin(180 - 2x) = CP*CP*sin(2x). Since the areas are equal, we conclude that CP = PD = AP = BP, so P is a circumcenter of ABCD, and the quadrilateral ABCD is cyclic, as desired.

See Also

1998 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

First, let's prove that if ABCD is cyclic, then the triangles ABP and CDP have equal areas. The circumcenter of ABCD is the intersection of perpendicular bisectors of the sides of the quadrilateral ABCD, so P is the circumcenter. From this, we derive that PC=PB=PA=PD. Since the diagonals AC and BD are perpendicular, the sum of the angles APB and CPD is equal to 180 degrees. Denote the angle APB by a. Then the angle CPD is equal to 180 - a. The area of the triangle APB is equal to AP*BP*sin(angle APB). Let R denote the radius of the circle. Then the area of the triangle APB is equal to R*R*sin(a), and the area of CPD equals to R*R*sin(180 - a) = R*R*sin(a), so the triangles ABP and CDP have equal areas. Now, let's prove it in the other direction. We know that ABP and CDP have equal areas. Let O denote the intersection of the diagonals AC and BD. Let M be the midpoint of AB and N be the midpoint of CD. Draw a segment between M and O, and draw another segment between O and N. We know that ON = CN = ND and OM = MB = AM. Since S(ABP) = S(CPD) (where S denotes the area of the triangles), we have AB*PM = CD*PN. From this, we can conclude that PM/PN = CD/AB = ON/OM. Let the intersection of the lines AB and CD be the point E. Since PM is perpendicular to AB and PN is perpendicular to CD, PMEN is a cyclic quadrilateral, so the angle MPN is equal to 180-(angle AED) = BAD + CDA (where BAD and CDA are angles). The angle MON is equal to BAC + CDB + 90 (since the triangles AMO and OND are isosceles). In triangle AOD, OAD+ODA = 90 (where OAD and ODA are angles), so the angle MON is equal to BAC + CDB + CAD + BDA = BAD + CDA = MPN. Now take a look at triangles MPN and NOM. We know that MP/PN = ON/OM and the angle MPN is equal to MON. This implies that the triangle MPN is similar to NOM. Since the two triangles have a common side, MN, we conclude that the triangles MPN and NOM are equal, so MP = ON = CN = CD/2, and PN = OM = MB = AB/2. From this, we can derive that the triangles MBP and NPD are equal, so the angle MPB = angle PDN. Denote the angle by x. In the same way, we can derive that the triangles MBP, MAP, NPC, and NPD are equal, so the sum of the angles APB and CPD is 180 degrees, and the angle APB is equal to 2x. So the area of ABP is equal to AP*AP*sin(2x) and the area of CPD is equal to CP*CP*sin(180 - 2x) = CP*CP*sin(2x). Since the areas are equal, we conclude that CP = PD = AP = BP, so P is a circumcenter of ABCD, and the quadrilateral ABCD is cyclic, as desired.