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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