Parallel Sections in a Trapezoid

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Your goal: Use similar triangles to locate parallel sections, distinguish two midpoint segments, and reconstruct a trapezoid from its section lengths.

Preparation context: Austria’s Junior Regional Competition (JRW) is a proof competition. Its current regional rules specify four problems, one each in algebra, geometry, combinatorics and number theory, with four hours and eight points per problem. The released 2026 JRW paper includes trapezoid and midpoint reasoning. This original lesson develops selected geometry foundations and proof bridges for that setting. It is not a full mock, a complete syllabus, an official problem or a claim of full JRW readiness. The optional multiple-choice checks test conclusions; the actual regional competition requires written solutions.

Study plan: Spend a first sitting on the diagnostic, section formula and Examples 1–2. Use a second sitting for the two midpoint segments and Examples 3–4. Then attempt the written problems over several sessions. Allow about 30–40 minutes per teaching sitting, and more time for independent proofs.

Prerequisites: Fractions and ratios, similar triangles, the triangle midpoint theorem, and simple equations. The final reconstruction uses the difference-of-squares identity, explained before it is needed.

Three prerequisite checks

Try these without a calculator. This unsaved starting check is separate from the adaptive practice session. Repair a missing prerequisite before continuing.

Check 1. A ratio is not yet a fraction of the whole

A point E lies inside segment AD and $AE:ED = 2:3$. What fraction of AD is AE?

Complete solution

The complete segment has $2 + 3 = 5$ equal ratio-parts. The first part AE uses two of them, so $\frac{AE}{AD}=\frac25$. The ratio $\frac{AE}{ED}=\frac23$ compares the two pieces; it is not the fraction of the complete segment. If this distinction was difficult, review fractions and ratios first.

Check 2. Similarity gives a length ratio

In triangle XYZ, U lies on XY and V on XZ, with UV parallel to YZ. If $\frac{XU}{XY}=\frac13$ and $YZ = 18$, find UV and explain the theorem used.

Complete solution

Parallel lines give matching angles, and the angle at X is shared. Thus triangles XUV and XYZ are similar by AA. Corresponding lengths have the same ratio: $\frac{UV}{YZ}=\frac{XU}{XY}=\frac13$. Therefore $UV = 18/3 = 6$. If you could calculate 6 but could not name the similar triangles, revisit the similar-triangles lesson.

Check 3. Recover two lengths from a sum and difference

Two positive numbers have sum 28 and difference 10. Find the larger and the smaller.

Complete solution

Call them $a > b > 0$. Adding $a + b = 28$ and $a - b = 10$ gives $2a = 38$, so $a = 19$. Subtracting gives $2b = 18$, so $b = 9$. Both checks work: $19 + 9 = 28$ and $19 - 9 = 10$. This will distinguish the sum and difference formulas for two different midpoint segments.

Cut the trapezoid into two familiar triangles

Throughout this lesson, ABCD is a convex trapezoid, with vertices named in order around its boundary, $AB\parallel CD$, and base lengths $AB = a > b = CD > 0$. Here trapezoid means a quadrilateral with parallel opposite bases; the strict inequality excludes a parallelogram. We do not assume that the legs are equal, that any angle is right, or that the vertices lie on a circle.

Choose E inside leg AD and draw the line through E parallel to the bases. It meets the other leg BC at F. Write $t=\frac{AE}{AD}$, so $0 < t < 1$. This measures the position from the longer base AB towards the shorter base CD. Our aim is to find $L(t) = EF$.

A trapezoid is not a triangle, so do not apply a triangle ratio to the whole shape. Draw diagonal AC and call its intersection with EF P. Now use the two triangles on the two sides of that diagonal.

1. In triangle ADC, $EP\parallel DC$. Triangles AEP and ADC are similar, so $\frac{EP}{b}=\frac{AP}{AC}=\frac{AE}{AD}=t$. Hence $EP = tb$.
2. Because $\frac{AP}{AC}=t$, the remaining fraction is $\frac{CP}{CA}=1-t$. In triangle CBA, $PF\parallel AB$, so triangles CPF and CAB are similar. Thus $\frac{PF}{a}=\frac{CF}{CB}=\frac{CP}{CA}=1-t$. Hence $PF = (1 - t)a$.
3. The point P lies between E and F because the diagonal lies inside the convex quadrilateral. Add the two positive pieces.

$L(t)=EF=(1-t)a+tb=a-t(a-b)$

The same similarity proves $\frac{BF}{BC}=t$: since $\frac{CF}{CB}=1-t$, the complementary fraction measured from B is t. So a parallel section cuts both legs at the same fraction from the longer base, even when the legs have different lengths.

Check the endpoints algebraically. At $t = 0$ the formula gives a, the length of AB; at $t = 1$ it gives b, the length of CD. For an interior section, $b < L(t) < a$, since a positive fraction of the difference $a - b$ is subtracted from a. These endpoint lines are useful checks, although our section points E and F are normally required to be interior.

Reverse the question: locate a given length

From $L = a - t(a - b)$, subtract L and divide by the positive difference $a - b$. The result is:

$t=\frac{a-L}{a-b}$

An interior section with prescribed length L exists exactly when $b < L < a$. Under that condition the displayed t lies strictly between 0 and 1; place E at that fraction of AD and draw the parallel. The formula also proves uniqueness: different t values give different lengths because $a - b > 0$.

Four connected worked examples

Example 1. Start with the fraction of a leg

The bases are 18 and 6. A parallel section is two fifths of the way from the longer base to the shorter one along either leg. Find its length.

Choose the method. The position fraction is supplied, so use the section formula with $a = 18$, $b = 6$ and $t=\frac25$.

$L=18-\frac25(18-6)=18-\frac{24}{5}=\frac{66}{5}$.

The length is $\frac{66}{5}$. It lies between 6 and 18, as an interior section must. Notice that it is closer to 18: the section has travelled less than halfway towards the shorter base. This is a check on orientation, not a substitute for the calculation.

Your turn 1. Change the fraction

The bases are 15 and 9. Find the parallel section one third of the way from the longer base.

Hint 1

Use the whole base difference, not only the longer base.

Hint 2

The difference is 6; one third of that difference is 2.

Complete solution

Use $a = 15$, $b = 9$, $t=\frac13$. Then $L=15-\frac13(15-9)=13$. The answer is between 9 and 15. Dividing the longer base by 3 alone would ignore the width remaining at the shorter base.

Example 2. A section length locates its line

The bases are 22 and 10. A section parallel to them has length 16. Prove that it joins the midpoints of the legs.

What changed? The length is known, while the position is hidden. Let the position fraction from AB be t. The section formula gives $16 = 22 - 12t$, so $12t = 6$ and $t=\frac12$.

Thus $\frac{AE}{AD}=\frac{BF}{BC}=\frac12$. E and F are the leg midpoints. This proves the conclusion; a diagram appearing halfway up would not prove it. The strict base inequality makes t unique.

Your turn 2. Convert the answer back into a ratio

The bases are 19 and 7. A parallel section has length 10. Find AE:ED, where E is its point on AD and A is on the longer base.

Hint 1

First find AE as a fraction of AD.

Hint 2

If AE uses three quarters of the whole leg, the remaining piece uses one quarter.

Complete solution

The fraction is $t=\frac{19-10}{19-7}=\frac34$. Therefore AE uses three quarters of AD and ED uses one quarter, giving $AE:ED = 3:1$. The value $3\!:\!4$ would compare AE with AD, not AE with ED.

Two midpoint segments: sum or difference?

Let E and F now be the leg midpoints. The section formula at $t=\frac12$ gives:

$EF=\frac{a+b}{2}$

Let P and Q be the diagonal midpoints, on AC and BD respectively. In triangle ADC, the midpoint theorem puts P on the parallel through E and gives $EP=\frac b2$. In triangle DAB, the midpoint theorem puts Q on the same line and gives $EQ=\frac a2$. Both P and Q lie on EF, and $a > b$ gives $EP < EQ$. The order is therefore E, P, Q, F.

$PQ=EQ-EP=\frac{a-b}{2}$

The full leg-midpoint section uses a sum; the shorter segment between the diagonal midpoints uses a difference. Name the endpoints before choosing the formula. Nothing in this proof required equal legs or perpendicular diagonals.

Example 3. Recover the bases from both midpoint segments

The segment joining the leg midpoints has length 12. The segment joining the diagonal midpoints has length 4. Find the two base lengths.

Call the longer base a and the shorter b. The sum formula gives $a + b = 24$, while the difference formula gives $a - b = 8$. Add these equations: $2a = 32$, so $a = 16$. Then $b = 24 - 16 = 8$.

Verify both pieces of information. The leg-midpoint length is $\frac{16+8}{2}=12$, and the diagonal-midpoint length is $\frac{16-8}{2}=4$. The positive ordered pair of bases is uniquely determined. These equations determine the base lengths, not the height or the angles of the trapezoid.

Your turn 3. Keep the two segments separate

The leg-midpoint segment has length 14 and the diagonal-midpoint segment has length 3. Find the bases.

Hint 1

One given length is half a sum; the other is half a difference.

Hint 2

The sum of the bases is 28 and their difference is 6.

Complete solution

The equations are $a + b = 28$ and $a - b = 6$. Adding gives $a = 17$; subtracting gives $b = 11$. Check: their half-sum is 14 and their half-difference is 3. The shorter base is not 3; that number described a segment between diagonal midpoints.

The section through the diagonal intersection

Let the diagonals AC and BD meet at O. Draw the parallel to the bases through O, meeting AD at E and BC at F. We shall show that O is the midpoint of EF. This is a different statement from saying that O is the midpoint of either diagonal, which is generally false.

Triangles AOB and COD are similar: the angles at O are vertically opposite, and the parallel bases give a second pair of equal angles. Corresponding sides give $\frac{AO}{OC}=\frac{AB}{CD}=\frac ab$. The whole diagonal AC is AO plus OC, so:

$\frac{AO}{AC}=\frac{a}{a+b}$

From the section proof, the fraction along AC agrees with the fraction along AD. Thus the section through O has $t=\frac{a}{a+b}$. Its two pieces are:

$EO=tb=\frac{ab}{a+b},\qquad OF=(1-t)a=\frac{ab}{a+b}$

The pieces are equal, proving the midpoint claim. Adding them gives the complete section length:

$H=EF=\frac{2ab}{a+b}$

This quantity is called the harmonic mean of the positive lengths a and b. You do not need a separate inequality theorem: we have derived the formula directly from similar triangles. Since $a > b$, its position fraction is greater than one half, so the intersection section lies on the shorter-base side of the leg-midpoint section.

Example 4. A midpoint of a section, not of a diagonal

The bases are 15 and 5. A line through the diagonal intersection O is parallel to the bases and meets the legs at E and F. Find EF, EO and the fraction AE/AD.

The position is $\frac{AE}{AD}=\frac{15}{15+5}=\frac34$. The section length is $EF=\frac{2\times15\times5}{15+5}=\frac{15}{2}$. O bisects this section, so $EO=\frac{15}{4}$.

The ordinary leg-midpoint section would have length $\frac{15+5}{2}=10$, so it cannot be the line just found. Likewise, $\frac{AO}{OC}=3$, so O does not bisect AC. The equal pieces are EO and OF only.

Your turn 4. Locate and check

For bases 12 and 6, find the section through the diagonal intersection, the length of each half, and its position fraction from the longer base.

Hint 1

Find the complete section before halving it.

Hint 2

Use 12 and 6 in the harmonic-section formula; the fraction along a leg is 12/18.

Complete solution

The position fraction is $\frac{12}{12+6}=\frac23$. The complete section is $\frac{2\times12\times6}{12+6}=8$, and each half has length 4. The answer is between 6 and 12 and below the leg-midpoint length 9, as expected.

Bridge to nonroutine problems: moving diagonal cuts

For a general interior parallel section EF, let $P=AC\cap EF$ and $Q=BD\cap EF$. The first section proof gives $EP = tb$. In triangle DAB, $EQ\parallel AB$. Triangles DEQ and DAB are similar, so $\frac{EQ}{a}=\frac{DE}{DA}=1-t$, giving $EQ = (1 - t)a$. Both distances are measured from E along the same segment EF, but their order can change. Therefore the distance between P and Q is:

$PQ=\left|(1-t)a-tb\right|=\left|a-t(a+b)\right|$

The vertical bars mean absolute value: take the nonnegative size of the difference. If $t<\frac{a}{a+b}$, then P comes before Q; if t is larger, Q comes before P. At equality the points coincide at O and the distance is zero. The midpoint picture alone cannot settle this order for every section.

A useful identity for reconstructing the bases

Write $M=\frac{a+b}{2}$ for the leg-midpoint length and $H=\frac{2ab}{a+b}$ for the intersection-section length. Put the difference over a common denominator:

$M-H=\frac{(a+b)^2-4ab}{2(a+b)}=\frac{(a-b)^2}{2(a+b)}$

The last step uses $(a+b)^2-4ab=a^2-2ab+b^2=(a-b)^2$. Our bases are positive and unequal, so the denominator is positive and the numerator is strictly positive. Thus $H < M$. Knowing M and H can recover the positive difference $a - b$, after which the sum-and-difference method finds both bases.

Common mistakes and a repair

• Measuring t from the wrong base. State $t=\frac{AE}{AD}$ with A on the longer base. Reversing direction changes t to $1 - t$.
• Treating a part-to-part ratio as a part-to-whole fraction. If $AE:ED = p:q$, then $t=\frac{p}{p+q}$.
• Using one triangle for the entire trapezoid. Draw a diagonal and add the two section pieces.
• Confusing midpoint endpoints. Leg midpoints give half the sum; diagonal midpoints give half the difference.
• Assuming a drawing is symmetric. Name the similar triangles and corresponding sides. Equal legs and right angles were never given.
• Keeping an old order of P and Q. Compare EP and EQ, or use an absolute value, before subtracting.
• Calling O the midpoint of every segment through it. It bisects the parallel section EF, but generally not either diagonal.
• Solving an equation but not checking the geometry. Verify $0 < t < 1$, positive bases, and $a > b$. Distinct values outside the permitted interval do not produce allowed sections.

Nine original problems with progressive hints

All problems use the convex trapezoid convention above: vertices A, B, C, D in boundary order, longer base AB of length a, shorter base CD of length b, and $a > b > 0$. A parallel section meets AD at E and BC at F; $t=\frac{AE}{AD}$. Lengths use the same unit. Write a justification before choosing an answer in the optional adaptive checks. Problems 7–9 are proof bridges; the objective check verifies only the requested numerical conclusion.

Foundation 1. Turn a ratio into a position

The bases are 25 and 10, and $AE:ED = 2:3$. Find EF.

Hint 1

Which segment is the denominator in the definition of t?

Hint 2

AE uses two out of five parts of AD. Subtract two fifths of the base difference from 25.

Complete solution

The whole leg has five ratio-parts, so $t=\frac25$. Hence $EF=25-\frac25(25-10)=19$. The value lies between the base lengths. The ratio $\frac23$ describes AE compared with ED and must not be substituted for t.

Self-check: State the correct fraction of the whole leg, use the parallel-section theorem, and check the resulting length lies between the bases.

Foundation 2. Complementary positions

The bases are 20 and 8. Find the sum of the lengths of the sections at $t=\frac14$ and $t=\frac34$. Then prove that, for any bases a and b and any $0 < t < 1$, the sections at t and $1 - t$ have sum $a + b$.

Hint 1

Compute each length from the same longer base.

Hint 2

For the proof, write the coefficients of a and b in the two section formulas and add them.

Complete solution

The base difference is 12. The two lengths are $20 - 12/4 = 17$ and $20 - 3 \times 12/4 = 11$, whose sum is 28.

For the general claim, write the two section lengths separately: $L(t) = (1 - t)a + tb$ and $L(1 - t) = ta + (1 - t)b$. Adding collects coefficient 1 for each base, giving $L(t) + L(1 - t) = a + b$. Both positions are interior because $0 < t < 1$. The result does not say that the two lengths are equal; they are equal only at the halfway position when the bases are unequal.

Self-check: Compute both numerical lengths, then prove the general sum by collecting both base coefficients.

Foundation 3. Two midpoint questions in one picture

The segment joining the leg midpoints has length 18, and the shorter base has length 12. Find the longer base and the distance between the diagonal midpoints.

Hint 1

Use the information about the leg midpoints to recover the missing base first.

Hint 2

The sum of the bases is 36; after finding the longer base, take half their difference.

Complete solution

The leg-midpoint formula gives $\frac{a+12}{2}=18$, so $a = 24$. The diagonal-midpoint distance is therefore $\frac{24-12}{2}=6$. These use different endpoint pairs: the first length is a half-sum, and the second is a half-difference. The bases 24 and 12 are positive and in the required order.

Self-check: Identify which midpoint formula applies at each step and verify the bases meet the given conditions.

Core 4. Recover the bases from two off-centre sections

The section one third of the way from AB to CD has length 18. The section two thirds of the way from AB to CD has length 14. Find both bases and prove they are uniquely determined.

Hint 1

Subtract the two section equations before trying to solve for both bases separately.

Hint 2

Moving from one third to two thirds removes another one third of the base difference, and the observed decrease is 4.

Complete solution

Let $d = a - b > 0$. The formula gives $a-\frac d3=18$ and $a-\frac{2d}{3}=14$. Subtract the second equation from the first: $\frac d3=4$, hence $d = 12$. Substitute into the first to obtain $a = 22$, then $b = a - d = 10$.

Checking gives $22 - 12/3 = 18$ and $22 - 2 \times 12/3 = 14$. Subtraction forced d, and substitution then forced a and b, so no other ordered pair of base lengths works. Such a convex trapezoid exists: for example, place bases of lengths 22 and 10 on two distinct parallel lines and join their corresponding endpoints in boundary order. No equal-leg assumption is needed.

Self-check: Use the same direction for both position fractions, solve a nondegenerate pair of equations, check both lengths, and explain uniqueness.

Core 5. When does a diagonal bisect a parallel section?

The bases are 24 and 8. A parallel section EF meets diagonal AC at P, and $EP = PF$. Find EF. Prove that P is the intersection O of the two diagonals.

Hint 1

Write the two pieces of EF using the diagonal-split proof.

Hint 2

The pieces are 8t and 24(1 − t). To prove P lies on BD, compare its distance from E with that of the cut on BD.

Complete solution

For a section at fraction t, similarity gives $EP = 8t$ and $PF = 24(1 - t)$. Equality therefore gives $8t = 24 - 24t$, so $32t = 24$ and $t=\frac34$. The section length is $24-\frac34(24-8)=12$.

Let Q be the point where the same line meets BD. The moving-cut formula gives $EQ = 24(1 - t)$, which equals EP. P and Q both lie on the ray from E towards F, so equal distances from E force $P = Q$. Thus P lies on both diagonals and is O. This justifies the intersection claim without assuming it from the picture.

Self-check: Solve the equality of section pieces, check the interior position, and justify the coincidence of the two diagonal cuts.

Core 6. The diagonal cuts have crossed

The bases are 21 and 9, and $t=\frac34$. The section EF meets AC at P and BD at Q. Find PQ, stating which of P and Q comes first when travelling from E to F.

Hint 1

Use two distances from E before subtracting.

Hint 2

Compare EP = 27/4 with EQ = 21/4. A distance is the larger minus the smaller.

Complete solution

Measure both distances from E. We have $EP=\frac34\times9=\frac{27}{4}$ and $EQ=\frac14\times21=\frac{21}{4}$. Since $\frac{21}{4}<\frac{27}{4}$, Q comes before P.

Consequently $PQ=EP-EQ=\frac64=\frac32$. The complete section has length $21-\frac34(21-9)=12$; both distances from E are between 0 and 12, consistent with the points lying on EF. Using $EQ - EP$ without checking the order would give a negative number, which cannot be a distance.

Self-check: Find both diagonal-cut positions, establish their order, and subtract in the correct direction.

Proof bridge 7. Find every section with a fixed diagonal gap

The bases are 18 and 6. Find all interior parallel sections EF for which the distance between the diagonal cuts P and Q is 2. Give each position fraction t and each length EF, and prove your list is complete.

Hint 1

The same positive distance can occur on either side of the diagonal intersection.

Hint 2

Set the signed difference 18 − 24t equal to both 2 and −2; retain only fractions strictly between 0 and 1.

Complete solution

The moving-cut formula gives $PQ=\left|18-24t\right|$. Therefore $PQ = 2$ is equivalent to exactly two possibilities: $18 - 24t = 2$ or $18 - 24t = -2$. Solving gives $t=\frac23$ or $t=\frac56$.

Both lie strictly between 0 and 1, so both produce interior sections. Their lengths are $18-12\times\frac23=10$ and $18-12\times\frac56=8$. Substitution into the absolute difference gives 2 in each case. The two signs exhaust the definition of absolute value; the linear equations have one solution each. Hence these are exactly the two required sections.

Self-check: Include both signs, check the allowed position interval, calculate each length, and explain why no other sign or root remains.

Proof bridge 8. Recover a scale from a length difference

The longer base is three times the shorter base. The leg-midpoint section is longer than the section through the diagonal intersection by 1. Find both bases and justify that no other positive pair has these properties.

Hint 1

Use one unknown for the shorter base and express all lengths in terms of it.

Hint 2

For bases 3b and b, the two section lengths are 2b and 3b/2.

Complete solution

Write the shorter base as $b > 0$ and the longer as $a = 3b$. The leg-midpoint length is $M=\frac{3b+b}{2}=2b$. The intersection-section length is $H=\frac{2\times3b\times b}{3b+b}=\frac{3b}{2}$; cancelling b is allowed because b is positive.

The given difference is $M-H=\frac b2=1$, so $b = 2$ and $a = 6$. Their section lengths are 4 and 3, with difference 1. The ratio condition left only the positive scale b, and the difference equation forced it to be 2, proving uniqueness.

Self-check: Use positivity when cancelling, distinguish the two means, verify the scale, and prove uniqueness.

Proof bridge 9. Two section lengths determine both bases

The leg-midpoint section has length 16, while the section through the diagonal intersection has length 12. Find the bases and the distance between the diagonal midpoints. Prove that the given lengths determine the base lengths uniquely.

Hint 1

The leg-midpoint section gives a sum. Use the difference between the two section lengths to recover a base difference.

Hint 2

The identity M − H = (a − b)²/[2(a + b)] gives (a − b)² = 256. Choose the sign consistent with a > b.

Complete solution

The first section gives $a + b = 32$. Use the identity proved in the bridge: $M-H=\frac{(a-b)^2}{2(a+b)}$. Substituting the two section lengths and the known sum gives $4=\frac{(a-b)^2}{64}$, hence $(a-b)^2=256$.

Because a is the longer base, $a - b > 0$; therefore $a - b = 16$. Adding and subtracting the sum and difference equations yields $a = 24$ and $b = 8$. The distance between the diagonal midpoints is $\frac{a-b}{2}=8$.

Check the original section lengths: $\frac{24+8}{2}=16$ and $\frac{2\times24\times8}{24+8}=12$. The sum was forced, the positive difference was forced, and together they force each base. Thus the base lengths are unique, although many heights and leg shapes remain possible.

Self-check: Use the derived identity, justify the positive square root, solve sum and difference, and verify both original lengths.

Write a complete geometry proof

This is an editorial self-check, not an official ÖMO marking scheme. The actual competition awards marks for written reasoning. No automatic check on this page assesses an arbitrary proof.

1. State the configuration. Name the parallel bases, longer base, point positions and any ratio direction.
2. Justify the tool. Name the similar triangles and their matching vertices, or the two midpoints required by the midpoint theorem.
3. Track the pieces. Decide whether to add lengths or subtract distances from the same point. Check point order before subtracting.
4. Explain the decisive algebra. Show how the geometry produces the equation; retain both signs when an absolute value requires them.
5. Finish the problem asked. Check positivity and interior positions. For “all” or “unique”, exclude unlisted cases. Verify the final values in the original section conditions.

Practise a conclusion, then explain it

The adaptive session selects six objective checks from the nine corresponding to the written tasks. It responds to your answers with hints, an explanation and links back to the relevant lesson section. Foundation checks ratios and midpoint distinctions; Core combines constructions; Stretch checks conclusions of the proof bridges. These labels describe progression within this lesson, not official contest grades.

A completed session means six responses, including requested solutions. It does not mean six independent correct answers, mastery, or a graded written proof. Use paper to supply the argument, compare it with the complete solution, and retry assisted work later. The existing optional browser-local progress and privacy choices apply; practice is free.

Interactive practice loads here. You can also use the complete question set below.

Return to the paper with a method

Without looking back, explain why a diagonal turns one trapezoid section into two triangle sections. Then sketch the two midpoint segments and state which one uses the base sum. Finally, explain why the diagonal intersection bisects a parallel section but generally does not bisect a diagonal. If any explanation is uncertain, use the linked theory before attempting a full proof paper.

Return to Austria’s study and official papers · Review similar triangles · Review triangle midpoints · Choose another lesson

Preparation sources and original material

Preparation context was checked against the official regional competition rules, the released 2026 Junior Regional Competition paper, and the 2027 competition calendar. Austria also has a parallel advanced route; the country guide explains the distinction. Current course contests are instructor-set and should not be confused with the standard regional format. Confirm eligibility, permitted aids and dates with the organiser.

All explanations, diagrams, worked examples and practice problems here are original IMOolympiad.com teaching material. Standard geometric facts are proved in the lesson. No official problem has been reproduced, translated or adapted by changing its numbers. Official papers are linked at their source. This independent educational site is not the organiser and claims no endorsement or affiliation. Context checked 6 November 2026.