HardMathematicsHigh School (11–12)Geometry › TrianglesNumericalConceptualOlympiad

In triangle ABC the sides are AB = 20, BC = 21 and CA = 13…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

In triangle $ABC$ the sides are $AB = 20$ , $BC = 21$ and $CA = 13$ . Check for yourself that the triangle is acute, so its altitudes meet at a point inside it. The altitude from $A$ meets $BC$ at $D$ , and the altitude from $B$ meets $CA$ at $E$ . These two altitudes cross at $H$ , the orthocentre of the triangle, and the third altitude also passes through $H$ . The circle through $A$ , $B$ and $C$ is the circumcircle, and its radius is $R$ . Reflect $H$ in the line $BC$ , and call the image $H'$ . Write $\Delta$ for the area of the triangle.

The figure shows all these points. It is drawn to scale, but your answers must come from reasoning, not from measuring.

Triangle ABC with B at the bottom left, C at the bottom right and A above, nearer to C. The sides are marked AB = 20, BC = 21 and CA = 13. The altitude from A meets BC at D and the altitude from B meets CA at E, with right-angle marks. The altitudes cross at H inside the triangle. The circumcircle passes through A, B and C. The line AD is continued below BC to the point H', the reflection of H in BC, which is drawn on the circle.

Find the lengths $BD$ and $AD$ .
Prove that $\angle BHD = \angle ACB$ . In the same way, $\angle CHD = \angle ABC$ . Deduce that $\angle BHC = 180^\circ - \angle BAC$ .
Prove that $H'$ lies on the circumcircle.
The line $AD$ meets the circumcircle again at $H'$ . Use the two chords $AH'$ and $BC$ , which cross at $D$ , to find $HD$ . Then find $AH$ .
Find $R$ , and check that $AH = 2R\cos A$ , where $A$ stands for the angle $BAC$ .

Give each length that is not a whole number to two decimal places. Each part may use the results of the earlier parts.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the foot $D$ by Pythagoras. Show that $H'$ is on the circumcircle by an angle chase, then use the crossing chords at $D$ to get $HD$ .

(a) From triangles $ABD$ and $ACD$ , $AD^2 = 20^2 - BD^2 = 13^2 - DC^2$ . So $BD^2 - DC^2 = 400 - 169 = 231$ . Since $BD + DC = 21$ , we get $BD - DC = 231 \div 21 = 11$ . Hence $BD = 16$ , $DC = 5$ and $AD = \sqrt{400 - 256} = 12$ . The triangle is acute because $13^2 + 20^2 > 21^2$ and the other two sums of squares are also larger than the third square, so $D$ lies between $B$ and $C$ .

(b) $E$ lies on $CA$ and $\angle BEC = 90^\circ$ , so in triangle $BEC$ we have $\angle EBC = 90^\circ - \angle C$ . The point $H$ is on $BE$ , so $\angle HBD = 90^\circ - \angle C$ . Triangle $BDH$ has a right angle at $D$ , so $\angle BHD = 90^\circ - \angle HBD = \angle C$ . The same argument with the altitude from $C$ gives $\angle CHD = \angle B$ . Since $D$ lies between $B$ and $C$ , $\angle BHC = \angle BHD + \angle DHC = \angle C + \angle B = 180^\circ - \angle A$ .

(c) A reflection in $BC$ keeps $B$ and $C$ fixed and keeps angles, so $\angle BH'C = \angle BHC = 180^\circ - \angle A$ . The point $H$ is inside the triangle, on the same side of $BC$ as $A$ , so $H'$ is on the other side. In the quadrilateral $ABH'C$ the opposite angles at $A$ and $H'$ add up to $180^\circ$ . So $ABH'C$ is cyclic, and $H'$ lies on the circumcircle.

(d) The line $AD$ is perpendicular to $BC$ , so the reflection of $H$ stays on it: $H'$ is on line $AD$ , below $BC$ , with $DH' = DH$ . The chords $AH'$ and $BC$ cross at $D$ , so $DA \cdot DH' = DB \cdot DC$ , that is, $12 \cdot DH' = 16 \times 5 = 80$ . So $HD = DH' = \dfrac{20}{3} \approx 6.67$ , and $AH = AD - HD = 12 - \dfrac{20}{3} = \dfrac{16}{3} \approx 5.33$ .

(e) The area is $\dfrac12 \times 21 \times 12 = 126$ . So $R = \dfrac{abc}{4\Delta} = \dfrac{20 \times 21 \times 13}{4 \times 126} = \dfrac{65}{6} \approx 10.83$ . By the cosine rule, $\cos A = \dfrac{13^2 + 20^2 - 21^2}{2 \times 13 \times 20} = \dfrac{128}{520} = \dfrac{16}{65}$ . Then $2R\cos A = \dfrac{65}{3} \times \dfrac{16}{65} = \dfrac{16}{3}$ , which is $AH$ .

Check. With $B = (0, 0)$ , $C = (21, 0)$ and $A = (16, 12)$ , the altitudes meet at $H = \left(16, \dfrac{20}{3}\right)$ , and the point $\left(16, -\dfrac{20}{3}\right)$ is at distance $\dfrac{65}{6}$ from the circumcentre, as (c) says.

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