Geometry problems in math olympiads

This is an ongoing survey of olympiad problems. Source material is INMO, USAMO and IMO. The goal is to indicate key ideas involved in the proof.

USAMO 2010 Problem 1

Let $A X Y Z B$ be a convex pentagon inscribed in a semicircle of diameter $A B$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $A X, B X$, $A Z, B Z$, respectively. Prove that the acute angle formed by lines $P Q$ and $R S$ is half the size of $\angle X O Z$, where $O$ is the midpoint of segment $A B$.

Key Concepts

  • Angle at centre is twice the angle at circumference
  • External angle in a cyclic quadrilateral is equal to the interior opposite angle
  • Simson Line
  • Cyclic Pentagon

Idea of Proof

If we use Simson line (feet of perpendiculars from a point in the circumference, in this case $Y$, to three sides of a triangle (in this case $\Delta A X B$ and $\Delta A Z B$) are collinear), then we immediately conclude that $PQ$ and $RS$ meet at a point $M$ on the diameter $AB$. Here is part of the picture.

Next notice that we have a cyclic pentagon $A P Y R M$. Now angle chasing shows $\angle Z M Y = \angle Z A Y$ and so on. This proof is suggested in Evan Chen’s notes.

Alternatively we may do angle chasing. Here the key idea is to show $M C Y D$ is a cyclic quadrilateral.

INMO 2014 Problem 5

In an acute-angled triangle $A B C$, a point $D$ lies on the segment $B C$. Let $O_1, O_2$ denote the circumcentres of triangles $A B D$ and $A C D$, respectively. Prove that the line joining the circumcentre of triangle $A B C$ and the orthocentre of triangle $O_1 O_2 D$ is parallel to $B C$.

Key Concepts

  • Similarity of triangles
  • Cyclic pentagon
  • Angles at orthocenter
  • Angle at the centre of a circle is twice the angle at circumference

Idea of Proof

Notice the $H$ and $O$ are orthocenter and circumcenter of two different triangles $\Delta O_1 O_2 D$ and $\Delta ABC$. One recurring theme in geometry is to bind special points in different triangles via cyclic pentagons. A major step in this problem is to identify $A O_1 H O O_2$ as a cyclic pentagon. In fact notice that $\Delta A O_1 O_2$ is similar to $\Delta A B C$ is similar to $O_1 O_2 D$.

INMO 2016 Problem 1

Let \( ABC \) be triangle in which \( AB = AC \). Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio \( \frac{AB}{BC} \).

Idea of proof

Suppose \( H \) is the orthocenter and \( I \) is the incenter. Suppose \(AD\) is the perpendicular cum angle bisector passing through \( H \) and \( I\) meeting \( BC \) at \( D \).

Then \( \tan \frac{B}{2} = \frac{r}{BD} \), the inradius.

We can also show \( \tan \frac{A}{2} = \frac{2r}{BD} \).

Since \( A = \pi – 2B \), doing some trigonometry we obtain the value of \( \cos B \). The final answer is equivalent to \( \frac{1}{2 \cos B } \) which turns out to be \( \frac {3}{4} \).

INMO 2014 Problem 1

In a triangle \( ABC \), let \(D\) be a point on the segment \(BC\) such that \(AB + BD = AC + CD\). Suppose that the points \(B, C\) and the centroids of triangles \(ABD\) and \(ACD\) lie on a circle. Prove that \(AB = AC\).

Key Concepts

  • Parallel lines divide sides in equal ratios
  • Apollonious Theorem
  • Triangular Inequality
  • Centroid divides median in \( \frac{2}{1} \) ratio.

Proof Idea

Suppose \( T \) is the midpoint of \(AD\). If \(G_1\) and \( G_2 \) are the two medians, we show that \(G_1 G_2\) is parallel to \( BC \).

Then \(B G_1 G_2 C\) is a cyclic trapezium. This implies \(BT = CT\)

Now apply Apollonious theorem to deduce that \(AB^2 + BD^2 = AC^2 + CD^2\)

Finally notice that \(AB + BD = AC + CD\) from given hypothesis. Transposing we get \(AB^2 – AC^2 = CD^2 – BD^2\). Hence either \(CD – BD\) cancels out with \(AB – AC \) provided both the non-zero, or we have \(AB + AC = BD + CD\). The second case is impossible as it violates triangular inequality. Hence the first case holds.


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