Plane Geometry

1.Similar triangles and the Pythagorean theorem

Two triangles are similar when their corresponding angles are equal. Then their corresponding sides are proportional: if triangle $XYZ$ is similar to $ABC$ with $X$ matching $A$, $Y$ matching $B$, and $Z$ matching $C$, there is a constant $k$, the scale factor, with $XY=k\,AB$, $YZ=k\,BC$, and $ZX=k\,CA$. Equal angles force proportional sides, so the AA, SAS and SSS criteria all reduce to this one idea. Write every ratio with the matching vertices in the same order. Most similarity mistakes are mismatched ratios, not false claims of similarity.

The most useful source of similar triangles is a line parallel to a side. Suppose a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$. The angles $\angle ADE$ and $\angle ABC$ are corresponding angles, and likewise $\angle AED=\angle ACB$. So triangles $ADE$ and $ABC$ are similar, with $\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$. The small triangle is a scaled copy of the big one, so every length on it is a fixed fraction of the matching length on the whole triangle. When a problem mentions a parallel line, write the scale factor down immediately.

Now the Pythagorean theorem. Let the right angle be at $C$, with legs $a=BC$ and $b=CA$ and hypotenuse $c=AB$. Drop the altitude from $C$ to $AB$, with foot $H$, and write $p=AH$ and $q=HB$, so $c=p+q$. Triangles $ACH$ and $ABC$ share the angle at $A$ and each has a right angle, so they are similar. Hence $\frac{AC}{AB}=\frac{AH}{AC}$, which is $b^2=pc$. In the same way $a^2=qc$. Adding gives $a^2+b^2=(p+q)c=c^2$. This proof is more useful than the bare formula, because the same similar triangles also give the altitude.

The altitude to the hypotenuse satisfies $h^2=pq$, where $h=CH$. The triangles $CHA$ and $CHB$ are similar to each other, so $\frac{CH}{HA}=\frac{HB}{CH}$. This gives $h=\sqrt{pq}$ directly from the two pieces of the hypotenuse, without computing the area. The converse of Pythagoras also holds. If $a^2+b^2=c^2$, build a right triangle with legs $a$ and $b$. Its hypotenuse is $c$ by Pythagoras, and the triangle with sides $a,b,c$ is congruent to it by SSS, so the original angle opposite $c$ is a right angle.

On the AMC the common right triangles are the $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$ and $20$-$21$-$29$ families, together with their multiples. Recognising a triple gives every side at once and saves a long computation. The practical habit is to test $a^2+b^2$ against $c^2$ as soon as two lengths are known. A triangle whose sides are $5,12,13$ is right even if the problem never says so.

• $a^2+b^2=c^2$ for a right triangle with legs $a,b$ and hypotenuse $c$
• Similar triangles have equal angles and proportional sides: $\frac{XY}{AB}=\frac{YZ}{BC}=\frac{ZX}{CA}$
• Altitude to the hypotenuse: $h^2=pq$, with $b^2=pc$ and $a^2=qc$

• Matching the sides in the wrong order when writing a similarity ratio.
• Applying Pythagoras when the angle is not known to be right.
• Treating the altitude to the hypotenuse as a leg.

2.Area: base-height, sine formula, and ratios of areas

Area is the first formula anyone learns, and on the AMC it is the most flexible tool for relating lengths. For a triangle with sides $a,b,c$ and altitudes $h_a,h_b,h_c$ to those sides, the area $K$ satisfies $K=\frac12ah_a=\frac12bh_b=\frac12ch_c$. You may choose any side as the base. When a problem gives one altitude, write the area in terms of that base, then reuse the same area for a different altitude. Since $ah_a=bh_b=2K$, the altitudes are inversely proportional to the sides.

The sine formula comes from the base-height formula. Take sides $b=AC$ and $c=AB$ meeting at $A$, with $AB$ as base. The altitude from $C$ to line $AB$ has length $b\sin A$, because it is the leg opposite $A$ in the right triangle formed by the altitude and $AC$. Therefore $K=\frac12 bc\sin A$, and by symmetry $K=\frac12ab\sin C=\frac12ca\sin B$. Whenever two sides and the included angle are known, this is the fastest route to the area. It also shows the area is largest when the included angle is $90^\circ$, since $\sin$ peaks there.

Ratios of areas are where contest time is saved. Two triangles with the same height have areas in the ratio of their bases, because $\frac12 b_1h$ and $\frac12 b_2h$ differ only by the factor $\frac{b_1}{b_2}$. Two triangles with the same base have areas in the ratio of their heights. The general identity is the one to memorise. If $D$ lies on line $AB$ and $E$ lies on line $AC$, then $\frac{[ADE]}{[ABC]}=\frac{AD}{AB}\cdot\frac{AE}{AC}$. Both triangles have the angle $A$, and the sine formula contributes exactly the two side ratios.

Apply the same reasoning to a cevian. If $D$ lies on side $BC$, then $\frac{[ABD]}{[ABC]}=\frac{BD}{BC}$, because both triangles share the altitude from $A$. The two pieces satisfy $[ABD]+[ADC]=[ABC]$, so their areas are proportional to the lengths of the parts of $BC$. Splitting a triangle along one side is a linear rule, and it is the basis of most mass-point and area-chasing arguments. The three medians of a triangle cut it into six triangles of equal area, a fact that follows by applying the base rule twice.

Common mistakes: using a base and an altitude that do not belong to the same side; dropping the factor $\frac12$ when writing the sine formula; and forming a ratio of two triangles that do not share a vertex, or whose opposite sides do not lie on a common line. When unsure, place the figure on coordinates and compute the shoelace area of each piece. The ratio is then a plain computation, and it cannot be fooled by a figure that looks symmetric but is not.

• $2K=a\,h_a=b\,h_b=c\,h_c$
• $K=\frac12 bc\sin A$
• $\frac{[ADE]}{[ABC]}=\frac{AD}{AB}\cdot\frac{AE}{AC}$ and $\frac{[ABD]}{[ABC]}=\frac{BD}{BC}$

• Using an altitude that is not perpendicular to the chosen base.
• Forgetting the factor $\frac12$ or the $\sin$ when moving from base-height to the sine formula.
• Taking a ratio of triangles that share no common vertex or height.

3.Heron's formula and the shoelace formula

Heron's formula gives the area of a triangle from its three sides alone, which is valuable when no altitude is easy to find. With semiperimeter $s=\frac{a+b+c}{2}$, the area is $K=\sqrt{s(s-a)(s-b)(s-c)}$. Read $s-a$ as the distance from the vertex $A$ to a tangency point of the incircle, and the formula becomes a statement about tangent lengths, which is one reason it appears alongside the incircle in competition problems.

Derivation. Start from $K=\frac12 bc\sin A$, so $16K^2=4b^2c^2\sin^2A=4b^2c^2-4b^2c^2\cos^2A$. By the law of cosines $2bc\cos A=b^2+c^2-a^2$, so $16K^2=4b^2c^2-(b^2+c^2-a^2)^2$. This difference of squares factors as $\bigl(2bc-(b^2+c^2-a^2)\bigr)\bigl(2bc+(b^2+c^2-a^2)\bigr)=\bigl(a^2-(b-c)^2\bigr)\bigl((b+c)^2-a^2\bigr)$. Each factor is again a difference of squares, giving $(a-b+c)(a+b-c)(b+c-a)(a+b+c)$. In terms of $s$ this is $16s(s-a)(s-b)(s-c)$, which is Heron's formula.

Example: the $13$-$14$-$15$ triangle has $s=21$, so $K=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84$. The product under the root is a perfect square, which is typical of triangles with integer area. Before applying the formula, check the triangle inequality, because Heron's formula silently returns a meaningless square root when three lengths do not form a triangle.

The shoelace formula handles triangles given by coordinates. For vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ listed in order around the triangle, $K=\frac12\left|x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_1-x_1y_3\right|$. The reason is that the area of triangle $OPQ$ with $O$ at the origin is $\frac12|x_Py_Q-x_Qy_P|$, which is the sine formula applied to the vectors $OP$ and $OQ$. The expression is invariant under translation, so a general triangle can be moved until one vertex sits at the origin.

The same cross-product idea gives the area of any polygon with vertices in order, by summing $\frac12(x_iy_{i+1}-x_{i+1}y_i)$ around the boundary. On the AMC, the shoelace formula is often the quickest way to finish a coordinate problem, because it avoids finding a foot of a perpendicular. Listing the vertices in consistent order keeps the signs right, and the absolute value removes any dependence on orientation.

Common mistakes: writing $s$ where $s-a$ belongs inside the root; forgetting that Heron's formula needs all three sides; and dropping the $\frac12$ in the shoelace formula. A useful check is that when $s$ is an integer and the four factors multiply to a perfect square, the area is an integer, as in the $13$-$14$-$15$ example.

• $K=\sqrt{s(s-a)(s-b)(s-c)}$ with $s=\frac{a+b+c}{2}$
• $16K^2=4b^2c^2-(b^2+c^2-a^2)^2$
• Shoelace: $K=\frac12\left|\sum (x_iy_{i+1}-x_{i+1}y_i)\right|$

• Using $s$ instead of $s-a$ inside the square root.
• Applying Heron's formula to three lengths that violate the triangle inequality.
• Omitting the factor $\frac12$ or the absolute value in the shoelace formula.

4.Medians, angle bisectors and altitudes

A median joins a vertex to the midpoint of the opposite side. Let $M$ be the midpoint of $BC$, so $BM=MC=\frac a2$, and let $m=AM$. Apply the law of cosines in triangles $ABM$ and $AMC$. The angles at $M$ are supplementary, so their cosines are opposite, and $c^2=m^2+\frac{a^2}{4}-am\cos\theta$, $b^2=m^2+\frac{a^2}{4}+am\cos\theta$. Adding the two equations gives the median formula $m^2=\frac{2b^2+2c^2-a^2}{4}$. It needs no angles, only the three sides.

The angle bisector from $A$ meets $BC$ at $D$, and the bisector theorem says $\frac{BD}{DC}=\frac{AB}{AC}=\frac cb$. Proof by areas: the bisector splits angle $A$ into two halves of $\frac A2$. Using the sine formula, $[ABD]=\frac12\,c\cdot AD\sin\frac A2$ and $[ADC]=\frac12\,b\cdot AD\sin\frac A2$, so $\frac{[ABD]}{[ADC]}=\frac cb$. Triangles $ABD$ and $ADC$ also share the altitude from $A$ to $BC$, so the same ratio equals $\frac{BD}{DC}$.

The length of the bisector follows from Stewart's theorem. For a cevian of length $d$ from $A$ to $BC$, splitting $BC$ into $m$ (next to $B$) and $n$ (next to $C$), Stewart's theorem reads $b^2m+c^2n=a(d^2+mn)$. For the bisector, $m=\frac{ac}{b+c}$ and $n=\frac{ab}{b+c}$. Then $b^2m+c^2n=abc$, and solving $abc=a(d^2+mn)$ gives $t_a^2=bc\left(1-\frac{a^2}{(b+c)^2}\right)$. Check the formula on the isosceles case $b=c$: it gives the altitude squared, as it should.

The altitude from $A$ has length $h_a=\frac{2K}{a}$. Two altitude facts appear constantly. The altitude to the hypotenuse of a right triangle is $\frac{ab}{c}$, and the distance from a vertex to the orthocenter is $AH=2R\cos A$, where $R$ is the circumradius. The second fact is derived in the circumcircle concept below. Keep the orthocenter in mind when a problem mentions three altitudes meeting at a point.

Common mistakes: writing the median formula with $+a^2$ instead of $-a^2$; reversing $b$ and $c$ in the bisector ratio, so that $BD$ is assigned the wrong part; and confusing the internal bisector with the external one. A quick sanity check is that a median is always shorter than the longer of the two sides adjacent to its vertex, and a bisector lies between the altitude and the median from the same vertex.

• $m_a^2=\frac{2b^2+2c^2-a^2}{4}$
• $\frac{BD}{DC}=\frac{AB}{AC}$ for the internal bisector
• $t_a^2=bc\left(1-\frac{a^2}{(b+c)^2}\right)$ and $h_a=\frac{2K}{a}$

• Sign error in the median formula: it is $2b^2+2c^2-a^2$, not $+a^2$.
• Reversing the bisector ratio, which swaps the two pieces of the opposite side.
• Confusing the internal and external bisector theorems.

5.Circle theorems: inscribed angles and tangents

The inscribed angle theorem says that the central angle subtending an arc is twice any inscribed angle subtending the same arc. Proof sketch: first take the case where one side of the inscribed angle is a diameter $CC'$. Triangle $OAC$ is isosceles with $OA=OC$, so its base angles are equal, and the exterior angle at $O$ is the sum of those two equal angles. This gives $\angle AOC'=2\angle ACC'$. For a general inscribed angle, draw the diameter through the vertex, split the angle into two pieces, and apply the diameter case to each.

Three consequences do most of the work. Inscribed angles on the same arc are equal. An angle inscribed in a semicircle is a right angle, which is why the hypotenuse of a right triangle inscribed in a circle is a diameter. The opposite angles of a cyclic quadrilateral sum to $180^\circ$, since they subtend arcs that together make the whole circle. The converse also holds, so showing that two opposite angles are supplementary proves a quadrilateral is cyclic.

Tangents bring in two more facts. The radius to a point of tangency is perpendicular to the tangent, so $OT\perp PT$ at the point of contact. Two tangent segments drawn from one external point have equal length, because the two right triangles formed with the radii are congruent by hypotenuse and leg. The tangent-chord angle is the angle between a tangent and a chord through the point of contact. It equals the inscribed angle on the arc cut off by the chord, which is half the arc.

A chord of length $\ell$ in a circle of radius $r$ lies at distance $d$ from the centre, where $\ell=2\sqrt{r^2-d^2}$. This is Pythagoras applied to the right triangle formed by the centre, the chord's midpoint and an endpoint of the chord. Use this relation whenever a problem gives a chord length and asks for a distance from the centre, or the reverse.

When two chords cross inside a circle, the angle between them equals half the sum of the two arcs they intercept, the arc inside the angle and its vertical partner. This generalises the inscribed angle theorem and explains why the intersecting chords and secants results are natural. Common mistakes: confusing an arc's measure with the angle that subtends it; using a central angle where the inscribed angle is needed; and assuming a tangent is perpendicular to a chord, which is true only when the chord is a diameter.

• Central angle $=2\times$ inscribed angle on the same arc
• An angle inscribed in a semicircle is $90^\circ$; opposite angles of a cyclic quadrilateral sum to $180^\circ$
• Tangent-chord angle $=\frac12$ (arc cut off), and $\ell=2\sqrt{r^2-d^2}$ for a chord at distance $d$

• Taking the inscribed angle to be the full arc rather than half of it.
• Assuming the tangent is perpendicular to a chord instead of to the radius at the point of contact.
• Using the central angle at the centre when the vertex lies on the circle.

6.Power of a point

Fix a circle with centre $O$ and radius $r$, and a point $P$. The power of $P$ is $p(P)=OP^2-r^2$. The theorem is that for every line through $P$ that meets the circle at points $X$ and $Y$, the product of the signed distances $\overline{PX}\cdot\overline{PY}$ is the same number $p(P)$, whatever the line. Proof: parametrise the line as $P+tu$ for a unit vector $u$. The points on the circle satisfy $|P+tu-O|^2=r^2$, which expands to $t^2+2t\,(P-O)\cdot u+(OP^2-r^2)=0$. The two roots $t_1,t_2$ give $X$ and $Y$, and by Vieta their product is $t_1t_2=OP^2-r^2$.

The tangent case is the same theorem with the two intersection points coinciding. If $PT$ is tangent at $T$, then $OT\perp PT$, so $PT^2=OP^2-r^2$ by Pythagoras. Thus the tangent length squared is the power, and any secant through $P$ satisfies $PX\cdot PY=PT^2$. This is the single most used form, and it turns a tangent length into a product of secant pieces with no angles at all.

For a point inside the circle, the power is negative, $p(P)=-PA\cdot PB$ with $PA\cdot PB$ positive, and the intersecting chords theorem is the statement that $AP\cdot PB=CP\cdot PD$ for any two chords through $P$. The proof through similar triangles is also instructive. Inscribed angles on the same arc make $\triangle PAC\sim\triangle PDB$, which gives $\frac{PA}{PD}=\frac{PC}{PB}$. Cross-multiplying yields the product identity.

Power of a point is valuable because it converts a configuration of lines and circles into a multiplicative equation with no angles. Whenever a problem contains a point together with two or more segments from it that end on a circle, write the product of the segments on each line and set them equal. The tangent is the degenerate case, and that is why it is often the first piece of information to extract.

Common mistakes: using the whole secant length instead of the external piece when the point is outside; forgetting that the external piece is the distance from $P$ to the nearer point; and mixing the chord products for a point inside with those for a point outside. The sign convention is harmless if you always use lengths, but it matters in coordinate proofs.

• $p(P)=OP^2-r^2$
• Tangent-secant: $PT^2=PX\cdot PY$
• Intersecting chords: $AP\cdot PB=CP\cdot PD$

• Using the far point of the secant where the near point is needed, or the reverse.
• Forgetting to square the tangent length.
• Applying the tangent formula when the point lies inside the circle.

7.Incircle and circumcircle: r and R

The incircle touches each side once. Join the incentre $I$ to the three vertices: the triangle splits into three pieces, each with height $r$ from $I$ to a side. Adding their areas gives $K=\frac12 r(a+b+c)=rs$, so $r=\frac Ks$. Combined with Heron's formula, $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$. This is the cleanest way to find the inradius from the sides, and it is why the semiperimeter appears so often in contest solutions.

Tangent lengths give the most useful coordinates. The two tangent segments from a vertex to the incircle are equal, because the two right triangles formed with $I$ and the tangency points are congruent. If the incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$, then $AE=AF=s-a$, $BD=BF=s-b$, and $CD=CE=s-c$. In coordinates, placing $BC$ on the $x$-axis and reading off $BD=s-b$ fixes the tangency point.

The circumradius satisfies $R=\frac{abc}{4K}$. The law of sines gives $a=2R\sin A$, and $K=\frac12 bc\sin A$. Multiplying $a=2R\sin A$ by $bc$ gives $abc=2R\cdot bc\sin A=4RK$. For a right triangle the circumcentre is the midpoint of the hypotenuse, so $R=\frac c2$. For an equilateral triangle of side $s$, $R=\frac{s}{\sqrt3}$ and $r=\frac{s}{2\sqrt3}$.

Two distances between centres are standard. Euler's formula $OI^2=R^2-2Rr$ gives the distance between the circumcentre $O$ and incentre $I$. It can be checked on the $13$-$14$-$15$ triangle, where $R=\frac{65}{8}$, $r=4$, and $OI^2=\frac{65}{8}\cdot\frac18=\frac{65}{64}$. The distance from the circumcentre to a side is $R\cos A$, since the central angle over that side is $2A$. The excentre opposite $A$ has radius $r_a=\frac{K}{s-a}$.

Common mistakes: using $s$ where $s-a$ is needed for a tangent length; using the semiperimeter in $R=\frac{abc}{4K}$ instead of the sides; and assuming the circumcentre lies inside an obtuse triangle. For an obtuse triangle $O$ lies on the other side of the longest side from the obtuse vertex, so check the angle before reading a distance off the formula $R\cos A$.

• $r=\frac Ks$ and $R=\frac{abc}{4K}$
• Tangent lengths from $A,B,C$ are $s-a,\ s-b,\ s-c$
• $OI^2=R^2-2Rr$ (Euler) and $r_a=\frac{K}{s-a}$

• Using $s$ where $s-a$ belongs in a tangent length.
• Confusing $r$ and $R$ in Euler's formula.
• Forgetting that the circumcentre of an obtuse triangle lies outside the triangle.

8.Cyclic quadrilaterals and Ptolemy

A cyclic quadrilateral has all four vertices on one circle. Its opposite angles are supplementary, which follows from the inscribed angle theorem: the two opposite angles subtend arcs that together make the full circle. The converse holds too. If $\angle A+\angle C=180^\circ$ in a convex quadrilateral, then $ABCD$ is cyclic. Recognising a cyclic quadrilateral lets you use the circle's theorems, especially power of a point and inscribed angles.

Ptolemy's theorem states that for a cyclic quadrilateral $ABCD$, $AC\cdot BD=AB\cdot CD+BC\cdot DA$. Proof: choose a point $E$ on the diagonal $BD$ with $\angle BAE=\angle CAD$. Triangles $ABE$ and $ACD$ have equal angles at $A$ and at $B$ and $C$ (inscribed angles on arc $AD$), so they are similar, giving $AB\cdot CD=AC\cdot BE$. Triangles $ABC$ and $AED$ are similar in the same way, giving $BC\cdot AD=AC\cdot ED$. Adding the two identities and using $BE+ED=BD$ gives the theorem.

Ptolemy becomes especially powerful in symmetric figures. For a rectangle with sides $a$ and $b$, the theorem reads $d^2=a^2+b^2$, which is Pythagoras again. For an isosceles trapezoid with parallel sides $p$ and $q$ and legs $\ell$, the two diagonals are equal, say $d$, and Ptolemy gives $d^2=pq+\ell^2$, because the pairs of opposite sides are $(p,q)$ and $(\ell,\ell)$. A diagonal of an isosceles trapezoid can therefore be found in one line once the bases and legs are known.

Brahmagupta's formula gives the area of a cyclic quadrilateral with sides $a,b,c,d$ as $K=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s=\frac{a+b+c+d}{2}$. It is Heron's formula with one side equal to zero, and it holds exactly when the quadrilateral is cyclic, because the cyclic condition makes the two opposite angles supplementary, which turns the area into a sum of two sine terms with equal sine.

Common mistakes: applying Ptolemy to a quadrilateral that is not cyclic; pairing sides incorrectly so that a side is multiplied by the wrong diagonal; and forgetting that Ptolemy relates products of two diagonals, not the diagonals individually. In any problem with four concyclic points, list the sides in cyclic order before you write the identity.

• Opposite angles of a cyclic quadrilateral sum to $180^\circ$
• Ptolemy: $AC\cdot BD=AB\cdot CD+BC\cdot DA$
• Brahmagupta: $K=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

• Applying Ptolemy when the quadrilateral is not cyclic.
• Listing the sides out of cyclic order, which swaps the pairs in the identity.
• Forgetting that Ptolemy's theorem gives a product of diagonals.

9.Special triangles

The $45^\circ$-$45^\circ$-$90^\circ$ triangle is half a square. If its legs are $a$, the hypotenuse is $a\sqrt2$, by Pythagoras with $a^2+a^2=c^2$. Its sides are in the ratio $1:1:\sqrt2$. Whenever a problem has a square, the diagonal of the square and the isosceles right triangles it creates give this ratio instantly, and a square diagonal is never a surprise.

The $30^\circ$-$60^\circ$-$90^\circ$ triangle is half an equilateral triangle. Take an equilateral triangle of side $2a$ and drop an altitude. It bisects the opposite side, so it splits the triangle into two copies of the $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $2a$ and short leg $a$. The long leg is $\sqrt{(2a)^2-a^2}=a\sqrt3$. The sides are in the ratio $1:\sqrt3:2$, with the short leg opposite $30^\circ$ and the long leg opposite $60^\circ$.

The equilateral triangle itself has a short list of formulas. With side $s$, the altitude is $\frac{\sqrt3}{2}s$, the area is $\frac{\sqrt3}{4}s^2$, the circumradius is $\frac{s}{\sqrt3}$, and the inradius is $\frac{s}{2\sqrt3}$. The inradius is one third of the altitude, since the centre divides each altitude in the ratio $2:1$. Keep these in mind so that they never need to be rederived under time pressure.

Pythagorean triples generate more special triangles. Every primitive triple has the form $m^2-n^2$, $2mn$, $m^2+n^2$ with $m>n$ of opposite parity and coprime. The triples $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$ and $7$-$24$-$25$ come from small choices of $m$ and $n$. A problem that gives one of these pairs of sides is often built around the triple, and the third side is then immediate.

Common mistakes: confusing which leg is opposite which angle in the $30$-$60$-$90$ triangle; writing the long leg as $2a$ instead of $a\sqrt3$; and applying the equilateral formulas to a triangle that is only isosceles. When a figure contains a $60^\circ$ angle and an altitude, look for the $30$-$60$-$90$ triangle hidden inside before reaching for trigonometry.

• $45^\circ$-$45^\circ$-$90^\circ$: sides $1:1:\sqrt2$
• $30^\circ$-$60^\circ$-$90^\circ$: sides $1:\sqrt3:2$
• Equilateral side $s$: altitude $\frac{\sqrt3}{2}s$, area $\frac{\sqrt3}{4}s^2$, $R=\frac{s}{\sqrt3}$

• Swapping the short and long legs of the $30$-$60$-$90$ triangle.
• Writing the long leg as $2a$ instead of $a\sqrt3$.
• Using equilateral-triangle formulas for a triangle that is only isosceles.

Example 1: Hypotenuse and one leg

A right triangle has hypotenuse $25$ and one leg of length $7$. What is the length of the other leg?

Example 2: Area from two sides and the included angle

In triangle $PQR$, $PQ=4$, $PR=6$, and the angle $\angle QPR$ equals $60^\circ$. What is the area of triangle $PQR$?

Example 3: Intersecting chords

Chords $AB$ and $CD$ of a circle meet at a point $P$ inside the circle. Suppose $AP=2$, $PB=9$, and $CP=3$. What is the length of chord $CD$?

Example 4: Inradius from the sides

What is the radius of the inscribed circle of the triangle with side lengths $5$, $6$, and $7$?

Example 5: Equilateral triangle on a circle

Equilateral triangle $ABC$ is inscribed in a circle. Point $P$ lies on the minor arc $BC$, the arc not containing $A$, with $PB=2$ and $PC=6$. What is $PA$?

Example 6: Distance from circumcentre to a side

Triangle $ABC$ has $BC=14$, $CA=13$, and $AB=15$. Let $O$ be the circumcentre. What is the distance from $O$ to line $BC$?