INMO Geometry Questions 2026 with Solutions

INMO Geometry Questions 2026 will help you understand the depth of various important INMO Geometry Topics such as Steiner's Line, Simson Line, Radical Axis, and Homothety. The Homi Bhabha Centre for Science Education (HBCSE) conducts the INMO to select the finest mathematical talent for the International Mathematical Olympiad (IMO) training camp. Geometry is often considered the most challenging yet rewarding part of the paper. To help aspirants excel, here is compiled a list of INMO Geometry Most Expected Questions based on the latest training camp modules and DPPs.

How to Prepare for INMO

INMO Geometry PDF Resources

To aid your preparation, we have categorized the Daily Practice Problems (DPP) from the INMO Training Camp. These documents contain the core problems and detailed proofs essential for the 2026 cycle.

INMO Geometry Important Topics

Based on the INMO Training Camp 2026 curriculum, the following topics are crucial. Focusing on these ensures a robust foundation for the Olympiad:

Triangle Centers: Properties of Orthocenter (H), Incenter (I), and Circumcenter (O).

Cyclic Quadrilaterals: Intersecting Chord Theorem, Ptolemy's Theorem, and Harmonic sets.

Line Geometry: Simson Line, Nine-point circle properties, and Radical Axes.

Advanced Lemmas: Power of a Point, Nesbitt’s Inequality in Geometry, and Cosine Law applications.

Transformation Geometry: Reflections, Rotations, and Homothety.

INMO Geometry Questions 2026 with Solutions

Below are some of the INMO Geometry important questions curated from the 2026 training camp practice sets.

Q1

Let H be the orthocenter of an acute triangle ABC and let D and E be the feet of the altitudes from vertices B and C, respectively. Line DE intersects the circumcircle of triangle ABC at points F and G. Denote by S the projection of H onto line DE. Prove that
EF + DS = DG + ES.

Answer:
Assume ∠B ≥ ∠C. Let HS = x and DE = y. Angle chasing yields ∠DHS = ∠B and ∠EHS = ∠C. Here,
x = 2R cos A cos B cos C and y = a cos A.

The difference DS − ES is calculated as
x(tan B − tan C) = 2R cos A sin(B − C).

This relationship establishes the required equality of the segments.

Q2

The diagonals AD, BE, CF of a convex hexagon ABCDEF intersect at point M. Triangles ABM, BCM, CDM, DEM, EFM, FAM are acute. Prove that the circumcenters of these triangles are concyclic if and only if the areas of quadrilaterals ABDE, BCEF, CDFA are equal.

Answer:
Let O₁, O₂, …, O₆ be the circumcenters. The distance O₁O₄ equals 2R, where R is the circumradius of the corresponding triangle. By analogy,
O₁O₄ = O₂O₅ = O₃O₆.

The circumcenters are concyclic if and only if these distances are equal, which occurs exactly when the specified quadrilateral areas are equal.

Q3

Consider an equiangular hexagon ABCDEF. Prove that
AC² + CE² + EA² = BD² + DF² + FB².

Answer:
For an equiangular hexagon with side lengths a, b, c, d, e, f, the lemma
a − d = e − b = c − f
holds.

Applying the Law of Cosines,
AC² = a² + b² − 2ab cos α.

Summing similar expressions for both sets of diagonals and substituting the lemma values proves the equality.

Q4

Let ABC be a triangle such that |AB| < |BC|, and let I be its incenter. Let M be the midpoint of segment AC, and N be the midpoint of arc AC containing B. Prove that
∠IMA = ∠INB.

Answer:
Let N′ be the second intersection of the circumcircle and line BI. It is the midpoint of the arc CA not containing B. Using properties of the incenter and arc midpoints, triangles IMA and INB share equal angles, which gives ∠IMA = ∠INB.

Q5

Let M be an interior point of triangle ABC. Lines AM, BM, CM intersect the circumcircles of triangles MBC, MCA, MAB at D, E, F, respectively. Prove
|AD|/|MD| + |BE|/|ME| + |CF|/|MF| ≥ 9/2.

Answer:
Rewrite each ratio using intersecting chords. Let
x = |AM|·|BC|,
y = |BM|·|CD|,
z = |CM|·|DB|.

The inequality reduces to Nesbitt’s inequality:
x/(y + z) + y/(z + x) + z/(x + y) ≥ 3/2.

Adding the three unit ratios gives the total ≥ 9/2.

Q6

Let AD be the altitude of an acute triangle ABC. Points E and F lie on line AD such that |DE| = |DF|. Circumcircles of triangles BEF and CEF meet sides at K, M and L, N, respectively. Prove that lines AD, KM, LN are concurrent.

Answer:
The circumcenter of triangle BEF lies on BC, so BK is a diameter and ∠BMK = 90°. Similarly, ∠LNC = 90°. These perpendicular relationships identify a common orthocenter, proving concurrency on line AD.

Q7

Let B′ and C′ be reflections of B and C across lines AC and AB, respectively. The circumcircles of triangles ABB′ and ACC′ intersect at A and P. Prove that the circumcenter O of triangle ABC lies on AP.

Answer:
Using reflection properties and cyclic quadrilaterals, angle chasing shows that ∠OPA equals ∠OAP, implying collinearity of A, P, O.

Q8

In triangle ABC with altitudes AD, BE, CF and orthocenter H, let EF ∩ AD = G. Let AK be the diameter of the circumcircle intersecting BC at M. Prove that GM ∥ HK.

Answer:
Points A, F, H, E lie on a circle. Similar triangles AGE and AMB give
AG/AE = AM/AB.

Using the converse of the intercept theorem, this ratio implies GM ∥ HK.

Q9

In acute triangle ABC (|AB| > |AC|), OQ is a diameter of the circumcircle of triangle BOC. A line through A parallel to BC meets CQ at M; a line through A parallel to CQ meets BC at N. If T = AQ ∩ MN, prove T lies on the circumcircle of triangle BOC.

Answer:
Let AQ meet the circle again at T′. Since OQ is a diameter, ∠OT′Q = 90°. Using midpoint and parallelogram arguments, T′ lies on MN, hence T = T′ and lies on the circle.

Q10

In acute triangle ABC (|AB| > |BC|), let A₁, C₁ be the feet of altitudes. Let D be the second intersection of the circumcircles of triangles ABC and A₁BC₁. If Z is the intersection of tangents at A and C, and X, Y are intersections of A₁C₁ with ZA, ZC, prove D lies on the circumcircle of triangle XYZ.

Answer:
Using tangent properties and power of a point, points X, Y, Z, D are shown to be concyclic.

Q11

Let I be the incenter of triangle ABC and D lie on AC such that |AB| = |DB|. The incircle of triangle BCD touches AC and BD at E and F. Prove EF passes through the midpoint of DI.

Answer:
Segment calculations show quadrilateral EDJI is a parallelogram. Diagonals bisect each other, hence EF passes through the midpoint of DI.

Q12

In quadrilateral ABCD,
∠DAB = 110°, ∠ABC = 50°, ∠BCD = 70°.

Let M, N be midpoints of AB, CD. Point P lies on MN such that
|AM| : |CN| = |MP| : |NP|
and |AP| = |CP|.

Find ∠APC.

Answer:
Using similarity and cyclic arguments, APCK is cyclic, giving
∠APC = 80°.

Q13

In triangle ABC, if
|BC| + |AC| = 2|AB|
and
∠BAC − ∠CBA = 90°,

find cos(∠ACB).

Answer:
Let a + b = 2c and A = 90° + B. Using the Law of Sines, the condition a + b = 2c transforms into
sin A + sin B = 2 sin C.

Substituting A = 90° + B and C = 90° − 2B, we obtain
sin(90° + B) + sin B = 2 sin(90° − 2B),
which simplifies to
cos B + sin B = 2 cos(2B).

This leads to the quadratic equation
4 cos² C + cos C − 3 = 0.

Factoring gives
(4 cos C − 3)(cos C + 1) = 0.

Since cos C ≠ −1, the result is
cos(∠ACB) = 3/4.

Q14

On side AC of triangle ABC, points D and E satisfy D between C and E. Points F and G are defined using circumcircle intersections and parallels through E. Prove that D, E, F, G are concyclic.

Answer:
Angle chasing and parallel line arguments show all four points lie on one circle.

Q15

Let ABC be a right triangle at C. Points A′, B′, C′ are the pedals of the centroid onto BC, CA, AB. Find the ratio
Area(A′B′C′) : Area(ABC).

Answer:
The area of the pedal triangle equals 2/9 of the area of triangle ABC.

Q16

In triangle ABC with AB = AC, point D lies on AB. The tangent at D to the circumcircle of BCD meets AC at E. A second tangent from E touches the circle at F. If G = BF ∩ CD and H = AG ∩ BC, prove
BH = 2HC.

Answer:
Harmonic bundles and perspectivity yield the required segment ratio.

Q17

Let K be the symmedian point of triangle ABC. Points A₁, B₁, C₁ are defined so that AB, AC, A₁K, BC are cyclic (and similarly for others). Prove A₁, B₁, C₁ are collinear.

Answer:
Applying Menelaus’ theorem confirms collinearity.

Q18

In acute triangle ABC with orthocenter H, point P lies on the nine-point circle. Circumcircles of AEHP and AFHP meet CH and BH at Q and R. Show line QR passes through a fixed point.

Answer:
Using power of point arguments, QR always passes through a fixed center related to triangle ABC.

Q19

Prove that in an equiangular hexagon, the difference between opposite sides is constant.

Answer:
In hexagon ABCDEF with all angles 120°,
a − d = e − b = c − f.

This follows by extending sides to form equilateral triangles.

Q20

State the condition for circumcenters of sub-triangles in a hexagon to be concyclic.

Answer:
They are concyclic if and only if areas of ABDE, BCEF, CDFA are equal.

Preparation Tips for INMO Geometry 2026

To succeed in the INMO Geometry section, candidates should follow a rigorous mathematical approach:

Master the Basics: Ensure a deep understanding of standard theorems (Ceva, Menelaus, Power of a Point) before moving to complex Olympiad lemmas.

Drafting Proofs: Practice writing full, formal proofs. In INMO, the method and logical flow are as important as the final result.

Visualization: Use diagrams to identify cyclic quadrilaterals and collinear points that may not be immediately obvious.

Daily Problem Solving: Solve at least 2-3 high-level geometry problems daily from previous INMO and IMO shortlists to build stamina and intuition.