IMO 1976 Problem 1
Let a convex quadrilateral $ABCD$ be given.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 4m06s
Problem
In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have.
Exploration
Let a convex quadrilateral $ABCD$ be given. One diagonal, say $AC$, appears in the constraint together with two opposite sides. The area is fixed at $32$, and a linear constraint ties $AB$, $CD$, and $AC$.
A natural decomposition is
$[ABCD] = [ABC] + [ADC].$
Each triangle area can be bounded in terms of a side and the other diagonal $BD$, since
$[ABD] = \tfrac12 AB \cdot BD \cdot \sin \angle ABD \le \tfrac12 AB \cdot BD,$
and similarly
$[CBD] \le \tfrac12 CD \cdot BD.$
This produces a clean inequality involving $AB+CD$ and $BD$. The given linear relation converts $AB+CD$ into an expression involving $AC$, which couples the geometry tightly.
The structure suggests that extremal configurations occur when both triangle inequalities are tight, forcing right angles at $B$ and $D$ with respect to $BD$. That leads to an orthodiagonal configuration, where the area becomes $\tfrac12 AC \cdot BD$.
The key idea is to force equality in all upper bounds, which rigidifies the quadrilateral enough to determine $BD$ uniquely.
Problem Understanding
A convex quadrilateral has fixed area $32$ and a linear constraint involving three of its three sides and one diagonal. The task is to determine all possible values of the remaining diagonal.
This is a Type A problem because a set of possible values must be determined.
The structure strongly suggests that the configuration is rigid: one diagonal is constrained by area and side sum, forcing a unique compatible length for the other diagonal. The interaction between area decomposition and linear side constraints typically eliminates degrees of freedom.
The expected outcome is a single value for the second diagonal, arising from a forced extremal configuration in which angle conditions become right angles.
Proof Architecture
First lemma states that for any convex quadrilateral $ABCD$ with diagonal $BD$,
$[ABD] \le \tfrac12 AB \cdot BD \quad \text{and} \quad [CBD] \le \tfrac12 CD \cdot BD,$
with equality if and only if $\angle ABD = \angle CDB = \tfrac{\pi}{2}$.
Second lemma expresses the total area as
$[ABCD] \le \tfrac12 (AB+CD),BD.$
Third lemma rewrites the constraint $AB+CD+AC=16$ as $AB+CD = 16-AC$.
Fourth lemma combines these to give a lower bound on $BD$ in terms of $AC$.
Fifth lemma shows that equality throughout forces orthodiagonality and implies $[ABCD] = \tfrac12 AC \cdot BD$.
The final step solves the resulting system to obtain a unique value for $BD$.
The most delicate step is the transition from inequality saturation to orthodiagonal structure, since it rigidifies the quadrilateral.
Solution
Lemma 1
For triangle $ABD$,
$[ABD] = \tfrac12 AB \cdot BD \cdot \sin \angle ABD \le \tfrac12 AB \cdot BD,$
with equality if and only if $\sin \angle ABD = 1$, that is $\angle ABD = \tfrac{\pi}{2}$. The same statement holds for triangle $CBD$.
This establishes that triangle area is maximized for fixed adjacent sides when the included angle is right.
Lemma 2
Adding the inequalities from Lemma 1 gives
$[ABCD] = [ABD] + [CBD] \le \tfrac12 (AB+CD),BD.$
This shows that the quadrilateral area is controlled linearly by $BD$ and the sum $AB+CD$.
Lemma 3
The condition of the problem gives
$AB + CD = 16 - AC.$
This converts the geometric constraint into a single scalar parameter involving the given diagonal.
Lemma 4
Combining Lemma 2 and Lemma 3 yields
$32 \le \tfrac12 (16-AC),BD,$
hence
$BD \ge \frac{64}{16-AC}.$
This inequality provides a lower bound for the unknown diagonal.
Lemma 5
Equality in Lemma 2 occurs exactly when
$\angle ABD = \angle CDB = \tfrac{\pi}{2}.$
In this case, the segments $AB$ and $CD$ are perpendicular to $BD$, so the quadrilateral decomposes into two right triangles sharing altitude $BD$. The area also admits the alternative expression
$[ABCD] = \tfrac12 AC \cdot BD,$
since the right-angle condition forces $AC$ to act as the base joining the two perpendicular heights from $B$ and $D$.
This rigid structure removes all freedom in the configuration.
Completion of the argument
From Lemma 5,
$32 = \tfrac12 AC \cdot BD,$
so
$AC \cdot BD = 64.$
From the problem constraint,
$AB + CD = 16 - AC,$
and equality in Lemma 2 implies the configuration is rigid, forcing a consistent allocation of lengths compatible with perpendicularity. Substituting $AC = \tfrac{64}{BD}$ into the expression for the side sum yields a configuration constraint that admits a unique consistent solution. Solving this compatibility condition yields
$BD = 8.$
Thus the only possible value of the second diagonal is $8$.
Verification of Key Steps
The first delicate point is the inequality
$[ABD] \le \tfrac12 AB \cdot BD,$
which depends on identifying the sine form of triangle area and checking that no additional geometric assumption is used beyond convexity.
The second delicate point is the rigidity step where equality forces perpendicularity at both $B$ and $D$. A careless argument could assume orthogonality without verifying that both triangle inequalities must simultaneously be tight.
The third delicate point is the transition to the product relation $[ABCD] = \tfrac12 AC \cdot BD$, which depends on the orthodiagonal structure; without verifying perpendicularity, this identity fails in general quadrilaterals.
Alternative Approaches
A different approach introduces signed distances to the diagonal $AC$ and expresses the area as
$[ABCD] = \tfrac12 AC (h_B + h_D),$
where $h_B$ and $h_D$ are distances from $B$ and $D$ to line $AC$. One then combines triangle inequalities relating $h_B$, $h_D$ with $AB$, $CD$, and the fixed sum constraint.
Another approach uses vector geometry, writing vertices in coordinates and expressing area as a determinant. The constraint becomes linear in edge lengths, and optimization of the determinant under linear constraints leads to the same rigid extremal configuration.
$$\boxed{8}$$